Đa thức đối xứng và ứng dụng trong đại số.pdf

I HC THI NGUYN

TRNG I HC KHOA HC

PHM VN TH

MT S TNH CHT CA A THC

I XNG V NG DNG

TRONG I S

LUN VN THC S TON HC

Chuyn ngnh : PHNG PHP TON S CP

M s : 60. 46. 40.

Ngi hng dn khoa hc: TS. NGUYN VN MINH

THI NGUYN 2012

S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

Mc lc

M u 3

1 Khi nim c bn v a thc i xng 5

1.1 a thc i xng hai bin . . . . . . . . . . . . . . . . . . 5

1.1.1 Cc khi nim c bn . . . . . . . . . . . . . . . . . 5

1.1.2 Tng ly tha v cng thc Waring . . . . . . . . . 6

1.1.3 Cc nh l v a thc i xng hai bin . . . . . . 9

1.2 a thc i xng ba bin . . . . . . . . . . . . . . . . . . . 11

1.2.1 Cc khi nim c bn . . . . . . . . . . . . . . . . . 11

1.2.2 Tng ly tha v tng nghch o . . . . . . . . . . 12

1.2.3 Qu o ca n thc . . . . . . . . . . . . . . . . . 14

1.2.4 Cc nh l ca a thc i xng ba bin . . . . . . 16

1.2.5 a thc phn i xng . . . . . . . . . . . . . . . . 19

2 ng dng tnh cht ca a thc i xng gii mt s

bi ton i s 21

2.1 Mt s bi tp tnh ton . . . . . . . . . . . . . . . . . . . 21

2.2 Phn tch a thc thnh nhn t . . . . . . . . . . . . . . . 24

2.3 Phng trnh i xng v phng trnh hi quy . . . . . . . 27

2.4 Gii h phng trnh . . . . . . . . . . . . . . . . . . . . . 33

2.4.1 H phng trnh i xng hai n v ng dng . . . . 33

2.4.2 H phng trnh i xng ba n . . . . . . . . . . . 37

2.5 Tm nghim nguyn ca cc phng trnh i xng . . . . . 42

2.6 Chng minh cc ng thc . . . . . . . . . . . . . . . . . . 44

2.7 Chng minh bt ng thc . . . . . . . . . . . . . . . . . . 50

3 a thc i xng n bin v ng dng 58

3.1 Cc khi nim . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.2 Biu din cc tng ly tha qua cc a thc i xng c s 60

3.3 Cc nh l ca a thc i xng nhiu bin . . . . . . . . 63

1S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

3.4 a thc phn i xng nhiu bin . . . . . . . . . . . . . . 66

3.5 Phng trnh v h phng trnh . . . . . . . . . . . . . . . 68

3.6 Chng minh ng thc. Phn tch a thc thnh nhn t . 72

Kt lun 79

Ti liu tham kho 80


M u

Cc bi ton i s lun chim mt v tr quan trng i vi ton ph

thng, cng l lnh vc m cc nh nghin cu sng to ra rt y

v hon thin. Tnh i xng trong i s l mt trong nhng phn quan

trng ca i s s cp, cng l bi ton quen thuc trong cc ti liu lin

quan n i s s cp, cc k thi hc sinh gii quc gia v quc t.

Trong qu trnh gii nhiu bi ton i s hoc dng trc tip hoc

dng gin tip mi nhn ra l bi ton lin quan n a thc i xng,

nu gii mi bi ton ny mt cch n l s gp khng t kh khn v

tnh hiu qu khng cao khi gii cc bi ton cng loi. Vic nm bt c

y khi nim v cc tnh cht c bn ca a thc i xng, thng qua

p dng gii mt s bi ton lin quan n a thc i xng l vn

c nhiu ngi quan tm.

Lun vn ny gii thiu cc khi nim, tnh cht ca a thc i xng

v cc ng dng c bn gii cc bi ton i s thng gp trong chng

trnh ton s cp. Lun vn "Mt s tnh cht ca a thc i xng v

ng dng trong i s" gm c phn m u, ba chng ni dung, kt

lun v ti liu tham kho.

Chng 1. Cc khi nin c bn v a thc i xng.

Trong chng ny tc gi trnh by cc khi nim, tnh cht ca a thc

i xng hai bin, ba bin. Mt ng gp nh c ngha trong chng

ny l H qu 1.1 ca cng thc Newton. Cng thc ny thng c s

dng trong cc bi ton tnh gi tr biu thc.

Chng 2. ng dng tnh cht ca a thc i xng gii mt s bi

ton i s.

Chng ny tc gi trnh by cc ng dng ca a thc i xng bng

cc v d minh ha c th. Cc ng dng ny rt ph bin trong cc ti

liu v i s trong chng trnh ton ph thng.


Chng 3. a thc i xng n bin v ng dng.

Chng ny tc gi trnh by cc kin thc ca a thc i xng n bin

v mt s ng dng ph bin thng gp.

Lun vn nghin cu mt phn rt nh ca i s v thu c mt s

kt qu nht nh. Tuy nhin, lun vn chc chn cn nhiu thiu xt, nn

rt mong c s gp ca cc thy c, cc bn ng nghip v c gi

quan tm n ni dung lun vn lun vn ca tc gi c hon thin

hn.

Lun vn c hon thnh ti trng i hc Khoa hc - i hc Thi

Nguyn di s hng dn ca TS. Nguyn Vn Minh. Tc gi xin by t

lng bit n su sc ti s quan tm ca thy, ti cc thy c trong Ban

Gim hiu, Phng o to v Khoa Ton - Tin trng i hc Khoa hc.

ng thi tc gi xin cm n ti S GD - T tnh Yn Bi, Ban Gim

hiu, cc bn ng nghip ti trng THPT Hong Vn Th huyn Lc

Yn - Yn Bi v gia nh to iu kin cho tc gi hc tp v hon

thnh bn lun vn ny.

Thi Nguyn, ngy 10 thng 06 nm 2012.

Tc gi

Phm Vn Th


Chng 1

Khi nim c bn v a thc ixng

1.1 a thc i xng hai bin

1.1.1 Cc khi nim c bn

nh ngha 1.1 (Theo [2]). Mt n thc f(x,y) ca cc bin c lp x,

y (trng hp chung nht c th l cc s phc) c hiu l hm s c

dng

f(x, y) = aklxkyl,

trong akl 6= 0 l mt s (hng s), k, l l nhng s nguyn khng m.S akl c gi l h s, cn k+l c gi l bc ca n thc f(x,y) v

c k hiu l

deg[f(x, y)] = deg[axkyl] = k + l.

Cc s k, l tng ng c gi l bc ca n thc i vi cc bin x, y.

Nh vy, bc ca n thc hai bin bng tng cc bc ca cc n thc

theo tng bin.

Chng hn: 3x4y2 v x2y l cc n thc theo x, y vi bc tng ng

bng 6 v 3.

nh ngha 1.2 (Theo [2]). Hai n thc ca cc bin x, y c gi l

ng dng (tng t), nu chng ch c h s khc nhau. Nh vy, hai n

thc c gi l ng dng, nu chng c dng: Axkyl, Bxkyl(A 6= B).nh ngha 1.3 (Theo [2]). Gi s Axkyl v Bxmyn l hai n thc ca

cc bin x, y. Ta ni rng n thc Axkyl tri hn n thc Bxmyn theo

th t ca cc bin x, y, nu k > m, hoc k = m v l > n.


Chng hn: n thc 3x4y2 tri hn n thc 3x2y7, cn n thc x4y5

tri hn n thc x4y3.

nh ngha 1.4 (Theo [2]). Mt hm s P(x,y) c gi l mt a thc

theo cc bin s x, y, nu n c th biu din c di dng tng ca hu

hn cc n thc. Nh vy, a thc P(x,y) theo cc bin s x, y l hm s

c dng

P (x, y) =

k+l

Nh vy

sk = 1sk1 2sk2. (1.1)Cng thc (1.1) c gi l cng thc Newton, n cho php tnh sk theo

sk1 v sk2.Vi m=1, m=2, nh l 1.1 ng v

s1 = x+ y = 1,

s2 = x2 + y2 = (x+ y)2 2xy = 21 22.

Gi s nh l ng cho m < k. Khi sk1 v sk2 ln lt l cc athc bc k-1, k-2 ca 1 v 2. Theo cng thc (1.1) ta suy ra sk l a

thc bc k ca 1 v 2. Theo nguyn l quy np ta c iu phi chng

minh.

H qu 1.1. Vi m > n, ta c

sm+n = sm.sn n2 .smn. (1.2)Tht vy,

sm+n = xm+n + ym+n = (xm + ym)(xn + yn) xnyn(xmn + ymn) =

sm.sn n2 .smnS dng cng thc (1.1) v cc biu thc ca s1, s2 chng minh trn,

ta nhn c cc biu thc sau

s1 = x+ y = 1,

s2 = 21 22,

s3 = 31 312,

s4 = 41 4212 + 222,

s5 = 51 5312 + 5122.

Vic tnh cc tng ly tha sk theo cng thc lp (1.1) khng c thun

tin v phi bit trc cc tng sk v sk1. i khi ta cn c biu thc skch ph thuc vo 1 v 2. Cng thc tng ng c tm ra nm 1779

bi nh ton hc ngi Anh E.Waring.

nh l 1.2 (Cng thc Waring (Theo [2])). Tng ly tha sk c biu

din qua cc a thc i xng c s 1 v 2 theo cng thc

1

ksk =

[k/2]m=0

(1)m (k m 1)!m! (k 2m)!

k2m1

m2 , (1.3)

trong [k/2] k hiu l phn nguyn ca k/2.


Chng minh. Ta chng minh cng thc (1.3) bng phng php quy np.

Vi k=1, k=2 cng thc tng ng c dng

s1 = 1,1

2s2 =

1

221 2.

Nh vy, vi k=1, k=2 cng thc (1.3) ng. Gi s cng thc Waring

ng cho s1, s2, ...., sk1. chng minh cng thc ng cho sk ta sdng cng thc (1.1). Ta c

1

ksk =

1

k[1sk1 2sk2] =

=k 1k

1.m=0

(1)m (k m 2)!m! (k 2m 1)!

k2m11

m2

k 1k

2.n

(1)n (k n 3)!n! (k 2n 2)!

k2n21

n2 =

=1

k

m

(1)m (k m 2)! (k 1)m! (k 2m 1)!

k2m1

m2

1k

n

(1)n (k n 3)! (k 2)n! (k 2n 2)!

k2n21

n+12

Trong tng th hai thay n+1 bi m. Khi hai tng c th kt hp thnh

mt nh sau:

1

ksk =

1

k

(1)m (k m 2)! (k 1)m! (k 2m 1)!

k2m1

m2

1k

m

(1)m1 (k m 2)! (k 2)(m 1)! (k 2m)!

k2m1

m2 =

1

k

m

(1)m (k m 2)![

k 1m! (k 2m 1)! +

k 2(m 1)! (k 2m)!

]k2m1

m2 .

S dng cng thc

1

(m 1)! =m

m!,

1

(k 2m 1)! =k 2m

(k 2m)! ,

ta c

(k 1)(k 2m)m!(k 2m)! +

(k 2)mm!(k 2m)! =

k(k m 1)m!(k 2m)! .

Cui cng, v

(k m 1).(k m 2)! = (k m 1)!nn ta c cng thc cn phi chng minh:


1ksk =

[k/2]m=0

(1)m (k m 1)!m! (k 2m)!

k2m1

m2 ,

Cng thc Waring cho biu thc ca sn = xn + yn theo

1 = x+ y, 2 = xy sau ys1 = 1;s2 =

21 22;

s3 = 31 312;

s4 = 41 4212 + 222;

s5 = 51 5312 + 5122;

s6 = 61 6412 + 92122 232;

s7 = 71 7512 + 143122 7132;

s8 = 81 8612 + 204122 162132 + 242;

s9 = 91 9712 + 275122 302132 + 9142;

s10 = 101 10812 + 356122 504132 + 252142 252;

.......................................................................

1.1.3 Cc nh l v a thc i xng hai bin

nh l 1.3 (Theo [2]). Mi a thc i xng P(x,y) ca cc bin x, y u

c th biu din c di dng a thc p(1, 2) theo cc bin 1 = x+ y

v 2 = xy, ngha l

P (x, y) = p(1, 2) (1.4)

Chng minh. Trc ht ta xt trng hp n thc, trong ly tha ca

x v y cng bc, ngha l n thc dng axkyk. Hin nhin l

axkyk = a(xy)k = ak2 .

Tip theo, xt n thc dng bxkyl(k 6= l). V a thc l i xng, nn cs hng dng bxlyk. xc nh, ta gi s k < l v xt tng hai n thc

trn

b(xkyl + xlyk) = bxkyk(xlk + ylk) = bk2slk.

Theo cng thc Waring slk l mt a thc ca cc bin 1, 2, nn nhthc ni trn l mt a thc ca 1, 2.

V mi a thc i xng l tng ca cc s hng dng axkyk v

b(xkyl + xlyk), nn mi a thc i xng u biu din c dng a

thc theo cc bin 1 v 2.


nh l 1.4 (Tnh duy nht (Theo [2])). Nu cc a thc (1, 2) v

(1, 2) khi thay 1 = x + y, 2 = xy cho ta cng mt a thc i xng

P(x,y), th chng phi trng nhau, ngha l (1, 2) (1, 2) .Chng minh. t (1, 2) = (1, 2) (1, 2). Khi theo gi thitta c:

(x+ y, xy) = (x+ y, xy) (x+ y, xy) = P (x, y) P (x, y) = 0.Ta s chng t rng (1, 2) 0. D thy rng, sau khi m ngoc thbiu thc

f (x, y) := (x+ y)k(xy)l

l mt a thc ca cc bin x, y v c s hng tri nht theo th t cc

bin x, y l xk+lyl.

Gi s (1, 2) c dng

(1, 2) =k,l

Aklk1

l2.

tm s hng tri nht, ta chn trong (1, 2) cc s hng c k+l l

ln nht. Tip theo, trong cc s hng ni trn, chn ra cc s hng vi

gi tr ln nht ca l. V d, nu

(1, 2) = 3412 42132 + 142 6122 + 1132 71 + 52 + 8

th s hng c chn s l 142.

Nh vy, gi s chn c n thc Am1 n2 . Khi , nu thay

1 = x+ y, 2 = xy, th s hng tri nht ca s l Axm+nyn. Tht vy,

gi s Bk1l2 l n thc ty khc vi Ax

m+nyn. Khi theo cch chn

c hoc m+n > l+l, hoc m+n = k+l, nhng n > l. Trong c hai trng

hp th Axm+nyn tri hn Bxk+lyl.

Vy chng t rng Axm+nyn l n thc tri nht ca (x + y, xy), nn

(x+ y, xy) 6= 0,x, y nu (1, 2) 6= 0. Vy, ta c (1, 2) 0.V d 1.1. Biu din a thc sau theo cc a thc i xng c s

f (x, y) = x5+3x3y2x3y3+2xy47x2y2+y5+3x2y35xy35x3y+2x4y= x3y3 7x2y2 + (x5 + y5)+ 3 (x3y2 + x2y3)+ 2 (xy4 + x4y)

5(xy3 + x3y

)=

= x3y37x2y2+(x5 + y5)+3x2y2 (x+ y)+2xy (y3 + x3)5xy (y2 + x2)= 32 722 + s5 + 3221 + 22s3 52s2 =

= 32722+(515312+5122)+3122+22(31312)52(2122)= 51 3312 5212 + 2122 32 + 322.


1.2 a thc i xng ba bin

1.2.1 Cc khi nim c bn

nh ngha 1.9 (Theo [2]). Mt n thc (x, y, z) ca cc bin x, y, z

c hiu l hm s c dng

(x, y, z) = aklmxkylzm,

trong k, l,m N c gi l bc ca cc bin x, y, z;s aklm R = R\ {0} c gi l h s ca n thc, cn s k+l+m gil bc ca n thc (x, y, z).

nh ngha 1.10 (Theo [2]). Mt hm s P(x,y,z) ca cc bin x, y, z

c gi l mt a thc, nu n c th c biu din dng tng hu hn

cc n thc:

P (x, y, z) =

k+l+mnaklmx

kylzm.

Bc ln nht ca cc n thc trong a thc c gi l bc ca a thc.

nh ngha 1.11 (Theo [2]). a thc P(x,y,z) c gi l i xng ca

cc bin x, y, z, nu n khng thay i vi mi hon v ca x, y, z, ngha

l

P (x, y, z) = P (x, z, y) = P (y, x, z) = P (y, z, x) = P (z, y, x) = P (z, x, y).

Chng hn cc a thc di y l nhng a thc i xng theo cc

bin x, y, z

x4 + y4 + z4 2x2y2 2x2z2 2y2z2;(x+ y)(x+ z)(y + z);

(x y)2(y z)2(z x)2.

nh ngha 1.12 (Theo [2]). a thc i xng P(x,y) c gi l thun

nht bc m, nu:

P (tx, ty, tz) = tmP (x, y, z),t 6= 0nh ngha 1.13 (Theo [2]). Cc a thc

1 = x+ y + z, 2 = xy + yz + zx, 3 = xyz,

c gi l cc a thc i xng c s ca cc bin x, y, z.


1.2.2 Tng ly tha v tng nghch o

nh ngha 1.14 (Theo [2]). Cc a thc sk = xk+yk+zk, (k = 0, 1, ...),

c gi l tng ly tha bc k ca cc bin x, y, z.

nh l 1.5 (Cng thc Newton (Theo [2])). Vi mi k Z, ta c hthc

sk = 1sk1 2sk2 + 3sk3 (1.5)Chng minh. Tht vy, ta c

1sk1 2sk2 + 3sk3 =

= (x+ y + z)(xk1 + yk1 + zk1)

(xy + xz + yz)(xk2 + yk2 + zk2) + xyz(xk3 + yk3 + zk3) =

= (xk + yk + zk + xyk1 + xk1y + xzk1 + xk1z + yzk1 + yk1z)

(xk1y + xyk1 + xk1z + xzk1 + yk1z + yzk1+

+xyzk2 + xyk2z + xk2yz) + (xk2yz + xyk2z + xyzk2) =

= xk + yk + zk = sk.

nh l 1.6 (Theo [2]). Mi tng ly tha sk = xk + yk + zk u c th

biu din c di dng mt a thc bc n theo cc bin 1, 2, 3.

Chng minh. Ta chng minh nh l bng phng php quy np. Ta c

s0 = 3, s1 = x+ y + z = 1s2 = x

2 + y2 + z2 = (x+ y + z)2 2(xy + yz + zx) = 21 22.Nh vy, nh l ng vi n = 0, n = 1, n = 2. Gi s nh l ng vi

n = k 1, n = k 2, n = k 3(k > 3). Khi , theo cng thc Newton,nh l cng ng vi n = k.

Cng thc (1.5) cho php biu din cc tng ly tha sk theo cc a

thc i xng c s 1, 2, 3, nu bit trc cng thc biu din ca

sk1, sk2. nh l sau cho ta cng thc biu din trc tip sk theo cc athc i xng c s 1, 2, 3.



din qua cc a thc i xng c s theo cng thc

1

ksk =

l+2m+3n=k

(1)klmn (l +m+ n 1)!l!m!n!

l1m2

n3 . (1.6)

Cng thc (1.6) c chng minh bng phng php quy np vi s tr

gip ca cng thc (1.5). Nh cng thc Waring chng ta c th tm c

cc cng thc sau

Biu thc ca sn = xn + yn + zn tnh theo 1, 2, 3.

s0 = 3;s1 = 1;s2 =

21 22;

s3 = 31 312 + 33;

s4 = 41 4212 + 222 + 413;

s5 = 51 5312 + 5122 + 5213 523;

s6 = 61 6412 + 92122 232 + 6313 12123 + 323;

s7 = 71 7512 + 143122 7132 + 7413 212123 + 7123 + 7223;

s8 = 81 8612 + 204122 162132 + 242 + 8513 323123+

+122123 + 241

223 8223;

s9 = 91 9712 + 275122 302132 + 9142 + 9613454123 + 5421223 + 183123 9323 271223 + 333;

s10 = 101 10812 + 356122 504132 + 252142 252 + 10713605123 + 10031223 + 254123 401323 6021223 + 10133 + 152223;

.......................................................................

nh ngha 1.15 (Theo [2]). Cc biu thc

sk = xk + yk + zk =1

xk+

1

yk+

1

zk, (k = 1, 2, ...)

c gi l tng nghch o ca cc bin x, y, z.

Do cng thc (1.5) ng vi k Z, nn nu trong cng thc thayk bi 3 k, ta c

sk =23s1k 1

3s2k +

1

3s3k (1.7)

S dng cng thc (1.7) c th tm c cc biu thc ca cc tng nghch

o theo cc a thc i xng c s. Chng hn:


s1 =23s0 1

3s1 +

1

3s2 =

23.3 1

3.1 +

1

3(21 22) =

23

;

s2 =23s1 1

3s0 +

1

3s1 =

23.23 13.3 +

1

31 =

22 21323

;

s3 =23s2 1

3s1 +

1

3s0 =

23.22 213

23 13.23

+1

3.3 =

=32 3123 + 323

33;

s4 =23s3 1

3s2 +

1

3s1 =

=23.32 3123 + 323

33 13.22 213

23+

1

3.23

=

=42 41223 + 4223 + 22123

43

1.2.3 Qu o ca n thc

nh ngha 1.16 (Theo [2]). a thc i xng vi s cc s hng ti

thiu, mt trong cc s hng ca n l n thc xkylzm c gi l qu

o ca n thc xkylzm v c k hiu l O(xkylzm).

R rng l tm qu o ca n thc xkylzm cn phi b sung vo n

thc tt c cc hon v ca x, y, z. Vi k 6= l 6= m, ta c:O(xkylzm) = xkylzm + xkymzl + xlykzm + xlymzk + xmykzl + xmylzk.

V d 1.2 (Theo [5]). Ta c

O(x5y2z) = x5y2z + x5yz2 + x2y5z + x2yz5 + xy5z2 + xy2z5;

O(x3y) = O(x3yz0) = x3y + xy3 + x3z + xz3 + y3z + yz3.

Nu trong n thc xkylzm c hai s m no bng nhau, chng hn

k = l 6= m, thO(xkykzm) = xkykzm + xkymzk + xmykzk

Chng hn: O(xyz5) = xyz5 + xy5z + x5yz,

O(xy) = xy + yz + zx,

O(x3y3) = x3y3 + x3z3 + y3z3.

Cc trng hp ring ca qu o:


O(x) = O(xy0z0) = x+ y + z = 1,

O(xy) = O(xyz0) = xy + yz + zx = 2,

O(xyz) = xyz = 3, O(xk) = O(xky0z0) = xk + yk + zk = sk, k N.

nh l 1.8 (Theo [2]). Qu o ca mi n thc biu din c di

dng a thc theo cc n thc i xng c s.

Chng minh. Trc ht ta c O(xk) = sk, nn theo nh l (1.6), O(xk)

biu din c theo cc a thc i xng c s.

Trng hp qu o c dng O(xkyl). Ta c cng thc

O(xkyl) = O(xk)O(xl)O(xk+l)(k 6= l). (1.8)Tht vy, ta c

O(xk)O(xl)O(xk+l) = (xk+yk+zk)(xl+yl+zl)(xk+l+yk+l+zk+l) =

= (xk+l + yk+l + zk+l) + (xkyl + xlyk + xkzl + xlzk + ykzl + ylzk)(xk+l + yk+l + zk+l) =

= xkyl + xlyk + xkzl + xlzk + ykzl + ylzk = O(xkyl).

Nu k = l th cng thc (1.8) c thay bi cng thc sau:

O(xkyk) =1

2[(O(xk))2 O(x2k)]. (1.9)

T (1.8) v (1.9), suy ra cc qu o O(xkyl) biu din c di dng

a thc theo cc bin 1, 2, 3.

Cui cng, nu n thc xkylzm ph thuc vo c ba bin x, y, z, ngha

l k 6= l 6= m 6= 0, th n thc xkylzm s chia ht cho ly tha vi s mno ca xyz. V vy trong a thc O(xkylzm) c th a ly tha vi

s m no ca xyz = 3 ra ngoi ngoc, khi trong ngoc ch l qu

o ph thuc vo s bin t hn ba. Do , qu o O(xkylzm) biu din

c di dng a thc ca 1, 2, 3.

Bng cch trn ta d dng nhn c cc cng thc sau:


Qu o O(xkyl) biu din dng a thc theo 1, 2, 3.O(xy) = 2;O(x2y) = 12 33;O(x3y) = 212 222 13;O(x2y2) = 22 213;O(x4y) = 312 3122 213 + 523;O(x3y2) = 1

22 2213 23;

O(x5y) = 412 42122 33 + 7123 + 232 323;O(x4y2) = 21

22 232 2313 + 4123 323;

O(x3y3) = 32 + 323 3123;

.......................................................................

S dng cc cng thc biu din ca tng nghch o theo cc a thc c

s, d dng tm c cc qu o O(xkyk). Tht vy, ta c

sk =1

xk+

1

yk+

1

zk=ykzk + xkzk + xkyk

xkykzk=O(xkyk)

k3.

Suy ra

O(xkyk) = k3sk;O(x2y2) = 23s2 =

22 213;

O(x3y3) = 33s3 = 32 3123 + 323;

O(x4y4) = 43s4 = 42 41223 + 4223 + 22123;

1.2.4 Cc nh l ca a thc i xng ba bin

nh l 1.9 (Theo [2]). Mi a thc i xng ba bin x, y, z u c th

biu din di dng a thc theo cc bin 1 = x+y+z, 2 = xy+yz+zx,

3 = xyz.

Chng minh. Gi s f(x, y, z) l a thc i xng v axkylzm l mt

trong cc s hng ca f(x, y, z). Do tnh i xng, cng vi s hng trn,

f(x, y, z) cha qu o O(xkylzm) vi tha s chung l a. Nh vy ta c

f(x, y, z) = a.O(xkylzm) + f1(x, y, z), (1.10)

trong f1(x, y, z) l a thc i xng no vi t s hng hn. i vi

f1(x, y, z) ta li c cng thc tng t nh cng thc (1.9). Theo mt s

hu hn bc ni trn, ta c th phn tch a thc f(x, y, z) thnh tng

cc qu o. Theo nh l (1.8), mi qu o li l mt a thc theo cc

a thc i xng c s, do mi a thc i xng c th biu din c

dng a thc theo cc a thc i xng c s.


nh l 1.10 (nh l duy nht (Theo [2])). Nu cc a thc (1, 2, 3)

v (1, 2, 3) khi 1 = x + y + z, 2 = xy + yz + zx, 3 = xyz cho ta

cng mt a thc i xng P(x,y,z), th chng phi trng nhau, ngha l

(1, 2, 3) (1, 2, 3) .Chng minh. thun tin ta t

t1 = 1, t2 = 2, t3 = 3; x1 = x, x2 = y, x3 = z.

(t1, t2, t3) = (t1, t2, t3) (t1, t2, t3).Theo gi thit ta c

(1, 2, 3) = P (x1, x2, x3) P (x1, x2, x3) = 0.1 = x1 + x2 + x3, 2 = x1x2 + x1x3 + x2x3, 3 = x1x2x3

Ta chng minh l a thc khng, ngha l ng nht bng khng. t

Q(x1, x2, x3) = (x1 +x2 +x3, x1x2 +x1x3 +x2x3, x1x2x3) = (1, 2, 3).

R rng Q(x1, x2, x3) l a thc i xng. Ta vit li (t1, t2, t3) dng

(t1, t2, t3) = 0(t1, t2) + 1(t1, t2)t3 + 2(t1, t2)t23 + ...+ m(t1, t2)t

m3

v k hiu 1, 2 l nhng a thc i xng c s ca cc bin x1, x2. D

thy rng

k(x1, x2, 0) = k(x1, x2) (k=1, 2), 3(x1, x2, 0) = 3(x1, x2) = 0.

Theo iu kin ca bi ton ta c

Q(x1, x2, x3) = 0(1, 2)+1(1, 2)3+2(1, 2)23+...+m(1, 2)

m3 =

0, x1, x2, x3.Khi th

R(x1, x2) := Q(x1, x2, 0) = 0(1, 2) = 0,x1, x2.V R(x1, x2) l a thc i xng hai bin, nn theo nh l tnh duy nht

ca a thc i xng hai bin (1.4) suy ra 0 ng nht bng khng. Nh

vy ta c

Q(x1, x2, x3) = 3[1(1, 2) + 2(1, 2)3 + ...+ m(1, 2)m13 ] = 0,

x1, x2, x3.v 3 6= 0, nn a thcQ(x1, x2, x3) = 1(1, 2) + 2(1, 2)3 + ...+ m(1, 2)

m13 = 0,

x1, x2, x3.


Lp lun tng t nh trn suy ra 1 ng nht bng khng. Tng t c

2, 3, ..., m l nhng a thc khng. Vy l a thc khng.

biu din mt a thc i xng qua cc a thc i xng c s, mt

cch tng qut, ta tin hnh theo cc bc nh trong chng minh nh l

(1.9). Tuy nhin, trong trng hp a thc l thun nht, ta c th dng

phng php " h s bt nh". C s ca phng php ny l mnh

sau.

Mnh 1.1 (Theo [2]). Cho fm(x, y, z) l mt a thc i xng thun

nht bc m. Khi fm(x, y, z) c biu din qua cc a thc i xng c

s theo cng thc

fm(x, y, z) =

i+2j+3k=m

aijki1

j2

k3 , (i, j, k N).

Mnh 1.1 c suy ra t cc nh l ca a thc i xng vi

1, 2, 3 ln lt c bc l 1, 2, 3 i vi cc bin x, y, z. Di y l mt

s trng hp ring ca mnh .

f1(x, y, z) = a11;

f2(x, y, z) = a121 + a22;

f3(x, y, z) = a131 + a212 + a33;

f4(x, y, z) = a141 + a2

212 + a3

22 + a413;

f5(x, y, z) = a151 + a2

312 + a31

22 + a4

213 + a523;

trong , ai(i = 1, 2, ...) l cc hng s c xc nh duy nht (theo nh

l 1.10) v tm cc h s ny, ta cho x, y, z nhn cc gi tr c th thch

hp no .

V d 1.3. Biu din a thc sau y theo cc a thc i xng c s

f(x, y, z) = x4 + y4 + z4 2x2y2 2x2z2 2y2z2.Li gii. Ta c

f(x, y, z) = O(x4) 2O(x2y2) == (41 4212 + 222 + 413) 2(22 213) = 41 4212.

V d 1.4. Biu din a thc sau y theo cc a thc i xng c s

f(x, y, z) = (x y)2(y z)2(z x)2.Li gii. Do f(x, y, z) l a thc thun nht bc 6, nn theo mnh (1.1)

ta c


f(x, y, z) = a161 + a2

412 + a3

21

22 + a4

32 + a5

313 + a6

23 + a7123.

Nhn xt rng, f(x, y, z) c bc cao nht i vi tng bin l 4, nn

a1 = a2 = 0. tm cc h s cn li, ta cho (x,y,z) ln lt cc gi tr

(0, 1,1), (0, 1, 1), (1, 1,2), (1, 1, 1), (1, 1, 1), ta tm c a3 = 1,a4 = 4, a5 = 4, a6 = 27, a7 = 18. Vy ta c kt qu

f(x, y, z) = 2122 432 4313 2723 + 18123.

1.2.5 a thc phn i xng

nh ngha 1.17 (Theo [2]). a thc phn i xng l a thc thay i

du khi thay i v tr ca hai bin bt k.

V d: Cc a thc x y v x4y2 y4x2 + x4y y4x+ x3y2 x2y3, lcc a thc phn i xng hai bin, cn a thc (x y)(x z)(y z) la thc phn i xng ba bin n gin.

nh l 1.11 (nh l Benzout (Theo [2])). Gi s f(t) l a thc bc

n > 1 . Khi s d trong php chia ca a thc cho t a bng f(a). athc f(t) chia ht cho t a khi v ch khi f(a) = 0 .Chng minh. Tht vy, thc hin php chia a thc f(t) cho ta, ta c

f(t) = g(t)(t a) + r(t).V t a c bc bng 1, nn a thc d r(t) c bc bng khng, ngha lr(t)=r l hng s. Trong ng thc trn cho t = a, ta c r = f(a). T

suy ra f(t) chia ht cho t a khi v ch khi f(a) = 0.nh l 1.12 (Theo [2]). Mi a thc phn i xng hai bin f(x, y) u

c dng:

f(x, y) = (x y)g(x, y), (1.11)trong g(x, y) l a thc i xng theo cc bin x, y.

Chng minh. Ta thy rng f(x, y) l a thc phn i xng th

f(x, x) = 0, v theo nh ngha ta c

f(x, y) = f(y, x).Trong ng thc trn t y = x, th f(x, x) = f(x, x), suy ra f(x, x) = 0.Ta k hiu Fy(x) = f(x, y) l a thc ch theo bin x (coi y l tham s).

Theo nhn xt trn, ta c Fy(y) = 0. Theo nh l Bezout, a thc Fy(x)

chia ht cho x y, do f(x, y) chia ht cho x y, ngha l c19S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

f(x, y) = (x y)g(x, y),trong g(x, y) l a thc no . Trong cng thc (1.11) i ch ca x

v y ta c

f(y, x) = (y x)g(y, x),V theo gi thit f(x, y) = f(y, x) v cng v x y = (y x) , nn tac:

f(x, y) = (x y)g(y, x), (1.12)T (1.11) v (1.12), suy ra g(x, y) l a thc i xng theo cc bin x,y.

nh l 1.13 (Theo [2]). Mi a thc phn i xng ba bin f(x, y, z)

u c dng

f(x, y, z) = (x y)(x z)(y z)g(x, y, z),trong g(x, y, z) l a thc i xng theo cc bin x, y, z.

nh l (1.13) c chng minh tng t nh l (1.12). Trong a thc

phn i xng, cc a thc x y v T = (x y)(x z)(y z) ng vaitr rt quan trng v c gi l cc a thc phn i xng n gin nht

tng ng i vi a thc phn i xng hai bin v ba bin.

i vi a thc phn i xng thun nht, ta c kt qu sau.

Mnh 1.2 (Theo [2]). Cho fm(x, y, z) l mt a thc i xng thun

nht bc m. Khi

f3(x, y, z) = aT (x, y, z),

f4(x, y, z) = aT (x, y, z)1,

f5(x, y, z) = T (x, y, z)(a21 + b2),

f6(x, y, z) = T (x, y, z)(a31 + b12 + c3), trong a, b, c l cc

hng s.

nh ngha 1.18 (Theo [2]). Bnh phng ca a thc phn i xng n

gin nht gi l bit thc.

Nh vy, trong trng hp hai bin, bit thc ca cc bin x, y l

(x, y) = (x y)2cn trong trng hp ba bin, th bit thc ca cc bin x, y, z l

(x, y) = T 2 = (x y)2(x z)2(y z)2.D thy rng : (x, y) = 21 42. Theo v d (1.3), ta c

(x, y, z) = 4313 + 2122 + 18123 432 2723, (1.13)trong 1, 2, 3 l cc a thc i xng c s.


Chng 2

ng dng tnh cht ca a thc ixng gii mt s bi ton i s

2.1 Mt s bi tp tnh ton

Bi 2.1. Cho x+1

x= a. Tnh M = x13 +

1

x13.

( thi HSG lp 8, tnh Thi Nguyn nm 1997 - 1998).

Li gii. S dng cng thc Waring ta tnh s13 = x13 +

1

x13theo 1 =

x+1

x= a;2 = x.

1

x= 1 c:

M = s13 = 13.6

m=0

(1)m (13m 1)!m! (13 2m)!

132m1

m2 =

= 131 13111 2 + 659122 1567132 + 1825142 913152 + 13162 == a13 13a11 + 65a9 156a7 + 182a5 91a3 + 13a.Bi 2.2. Tm hai ch s tn cng ca phn nguyn s (

29 +

21)2010.

Li gii. t

a = (

29 +

21)2 = 50 + 2

609,

b = (

2921)2 = 50 2609,sn = a

n + bn.

Ta c s1 = 1 = 100, 2 = ab = 64. Theo cng thc Newton th

sn+2 = 1sn+1 2snhay

sn+2 = 100sn+1 64sn


Do s1 = 100 nn sn+2 chia ht cho 100, vi n l 3,5,7,... (n l s l). Suy

ra s1005 = (

29 +

21)2010 + (

2921)2010 chia ht cho 100.Do 0 < (

2921)2010 < 1, nn 2 ch s tn cng ca phn nguyn ca

s (

29 +

21)2010 l 99.

Bi 2.3. Cho dy s Un = (3 +

5

2)n + (

352

)n 2 (n = 1, 2, 3, ..., n)1. Tnh 5 s hng u ca dy.

2. Lp cng thc tnh Un+2 theo Un+1 v Un.

Li gii. t a = (3 +

5

2); b = (

352

); sn = an + bn.

Ta c s1 = 1 = a+ b = 3;2 = a.b = 1; s2 = 21 22.

p dng cng thc Newtn sn+2 = 1sn+1 2sn. Ta cs3 = 1s2 2s1 = 3.7 1.3 = 18s4 = 1s3 2s2 = 3.18 1.17 = 37s5 = 1s4 2s3 = 3.37 1.18 = 93

M Un = sn 2 nn U1 = 1;U2 = 5;U3 = 16;U4 = 35;U5 = 91. Dosn = Un+2 nn bng cch thay vo cng thc Newtn sn+2 = 1sn+12snta c

Un+2 + 2 = 3(Un+1 + 2) 1.(Un + 2)hay

Un+2 = 3Un+1 Un + 2Bi 2.4. Cho cc s thc dng a, b tha mn: a100 + b100 = a101 + b101 =

a102 + b102. Tnh a2008 + b2008.

Li gii. t s1 = 1 = a + b = 3;2 = a.b; sn = an + bn. S dng cng

thc Newtn ta c s102 = 1.s101 2.s100 m s102 = s101 = s100 nn1 2 = 1 hay a+ b ab 1 = 0 (a 1)(b 1) = 0 a = b = 1 do s2008 = a

2008 + b2008 = 2.

Bi 2.5. Cho x1, x2 l nghim ca phng trnh x2 2x 2 = 0. Tnh

x71 + x72.

Li gii. t sn = xn1 +x

n2 , ta c s1 = 1 = 2;2 = 2; s2 = 2122 = 8,

nn theo cng thc Newtn sn+2 = 1sn+1 2sn ta c s3 = 20; s4 = 56T h qu ca cng thc Newtn sm+n = smsn n2 smn vi m > n cs7 = s4s3 32s1 = 56.20 (2)3.2 = 1136.Bi 2.6 (Vit Nam, 1975 (Theo [2])). Khng gii phng trnh x3x+1 =0, hy tnh tng cc ly tha bc tm ca cc nghim.


Li gii. Gi x1, x2, x3 l cc nghim ca phng trnh cho. Theo cng

thc Vite, ta c 1 = x1 + x2 + x3 = 0, 2 = x1x2 + x2x3 + x1x3 =

1, 3 = x1x2x3 = 1. S dng cng thc Waring, ta cx81 + x

82 + x

83 = s8 = s8 =

81 8612 + 204122 162132 + 242 + 8513

323123 + 122123 + 241223 8223 = 10.

Bi 2.7. Tm a thc bc 7 c h s nguyn nhn x = 7

3

5+ 7

5

3lm

nghim.

Li gii. t a = 7

3

5; b = 7

5

3.

t sn = an + bn, ta c s1 = 1 = a+ b = x, 2 = ab = 1.

Theo cng thc Waring c

s7 = 71 7512 + 143122 7132

35

+5

3= x7 7x5 + 14x3 7x

15x7 105x5 + 210x3 105x 34 = 0.Vy a thc bc 7 cn tm l 15x7 105x5 + 210x3 105x 34.Bi 2.8. Cho x1, x2 l nghim ca phng trnh x

2 6x + 1 = 0.Chngminh rng sn = x

n1 + x

n2 , n N l s nguyn khng chia ht cho 5.

Li gii.

Ta chng minh sn Z bng phng php quy np.Vi n=0 c s0 = 2 ZVi n=1 c s1 = 6 ZGi s sk, sk+1 Z, (k N) ta cn chng minh sk+2 Z.Tht vy, do s1 = 1 = 6, 2 = 1 m sk+2 = 1sk+1 2skhay sk+2 = 6sk+1 sk. Vy sk+2 Z.T kt qu sk+2 = 6sk+1 sk m

sk+1 = 6sk sk1nn

sk+2 = 6(6sk sk1) sk = 35sk 5sk1 sk1,do sk+2 chia ht cho 5 khi v ch khi sk1 chia ht cho 5 m s0 = 2;s1 = 6; s2 = 34 khng chia ht cho 5 nn sn khng chia ht cho 5.


2.2 Phn tch a thc thnh nhn t

Lun vn trnh by hai phng php phn tch a thc i xng thnh

nhn t. Phng php th nht ta biu din a thc cho theo cc a

thc i xng c s 1, 2. Phng php th hai l phng php h s bt

nh.

Cc Bi tp trong mc ny c trch dn t [5].

Bi 2.9. Phn tch a thc sau thnh nhn t

f(x, y) = 2x4 + 7x3y + 9x2y2 + 7xy3 + 2y4

Li gii.Ta c

f(x, y) = 2(x4 + y4) + 7xy(x2 + y2) + 9x2y2 = 2s4 + 71s2 + 922

Thay s2 = 21 22, s4 = 41 4212 + 222 vo biu thc trn ta c

f(x, y) = 241 212 22a thc trn c bc hai i vi 2 v c cc nghim l: 2 = 221, 2 = 21,do

f(x, y) = (221 2)(21 2) = (x2 xy + y2)(2x2 + 3xy + y2).Bi 2.10. Phn tch a thc sau thnh nhn t

f(x, y) = 3x4 8x3y + 14x2y2 8xy3 + 3y4.Li gii. Ta c

f(x, y) = 3(x4 + y4) 8xy(x2 + y2) + 14x2y2 = 3s4 82s2 + 1422Thay s2 =

21 22, s4 = 41 4212 + 222 vo biu thc trn ta c

f(x, y) = 341 20212 + 3622y l mt a thc bc hai theo 2 v khng c nghim (nghim thc). V

vy ta khng th phn tch a thc thnh tch hai nh thc theo 2, ta vn

dng phng php h s bt nh th biu din a thc cho dng

f(x, y) = 3x4 8x3y + 14x2y2 8xy3 + 3y4

= (Ax2 +Bxy + Cy2)(Cx2 +Bxy + Ay2) (2.1)

ng thc (2.1) tha mn vi mi x, y, nn ta s tm cc h s A, B, C

bng phng php h s bt nh nh sau:

Vi x=y=1, ta c


4 = (A+B + C)2

suy ra

A+B + C = 2Nhn xt rng cc h s A, B, C c xc nh chnh xc n du ca

chng, v nu thay i du ca tt c cc s ny thnh ngc li th ng

thc (2.1) khng thay i. V vy, khng lm mt tnh tng qut, ta c

A+B + C = 2

Tip theo, vi x = 1, y = 1, ta c36 = (AB + C)2 suy ra A+B + C = 6.

Tip theo, vi x = 0, y = 1 , ta c AC = 3

Vy xc nh cc h s A, B, C ta gii cc h phng trnh{A+B + C = 2AB + C = 6AC = 3

v

{A+B + C = 2AB + C = 6AC = 3

H phng trnh th nht cho ta nghim A = 1, B = 2, C = 3. H thhai v nghim (nghim thc). Vy ta c kt qu

f(x, y) = 3x4 8x3y + 14x2y2 8xy3 + 3y4 == (x2 2xy + 3y2)(3x2 2xy + y2).

Tip sau y, ta t 1, 2, 3 l cc a thc i xng c s ca b ba

s x, y, z hoc a, b, c. Chng hn, vi b ba s x, y, z ta c 1 = x+ y+ z,

2 = xy + xz + yz, 3 = xyz.


f(x, y, z) = (x+ y)(y + z)(z + x) + xyz

Li gii. Ta c

f(x, y, z) = (1 z)(1 x)(1 y) + 3 =

= 31 21(x+ y + z) + 1(xy + yz + zx) 3 + 3 =

= 31 31 + 12 3 + 3 = 12 =

= (x+ y + z)(xy + yz + zx).



f(a, b, c) = (a+ b+ c)3 (b+ c a)3 (c+ a b)3 (a+ b c)3

Li gii. Ta c

f(a, b, c) = 31 (1 2a)3 (1 2b)3 (1 2c)3 =

= 231 + 621(a+ b+ c) 121(a2 + b2 + c2) + 8(a3 + b3 + c3) =

= 231 + 631 121s2 + 8s3 =

= 231 + 631 121(21 22) + 8(31 312 + 33) =

= 243 = 24abc

Vy

f(a, b, c) = (a+ b+ c)3 (b+ ca)3 (c+a b)3 (a+ b c)3 = 24abc.Bi 2.13. Phn tch a thc sau thnh nhn t

f(x, y, z) = 2x2y2 + 2y2z2 + 2z2x2 x4 y4 z4.Li gii. Ta c

f(x, y, z) = 2O(x2y2)s4 = 2(22213)(414212 +222 +413) =

= 41 + 4212 813 = 1(31 + 412 83)

Ta thy, a thc trn l a thc bc bn i vi 1 v chia ht cho 1. Hn

na, a thc cho l hm chn i vi x, y, z nn khi ta thay x bi -x

( hoc y bi -y, hoc z bi -z) th a thc khng thay i, nhng x+y+z tr

thnh -x+y+z (hoc x-y+z hoc x+y-z), do a thc cng chia ht cho

-x+y+z , x-y+z , x+y-z. Cng v a thc cho bc bn, nn ta c:

f(x, y, z) = (x+ y + z)(x+ y + z)(x y + z)(x+ y z).Ptrong P l hng s no .

Hng s P c xc nh bng phng php h s bt nh, bng cch cho

x = y = z = 1, ta c 3 = 3P hay P=1.

Vy ta c kt qu:

f(x, y, z) = (x+ y + z)(x+ y + z)(x y + z)(x+ y z).


2.3 Phng trnh i xng v phng trnh hi quy

a thc i xng l cng c hu hiu gii cc phng trnh i s

bc cao, c bit l phng trnh h s i xng v phng trnh hi quy.

nh ngha 2.1 (Theo [2]). a thc

f(z) = a0zn + a1z

n1 + ...+ an; (a0 6= 0)c gi l a thc i xng, nu cc h s cch u hai u bng nhau,

ngha l

a0 = an, a1 = an1, a2 = an2, ...

Phng trnh ca a thc i xng c gi l phng trnh i xng.

Chng hn, cc a thc sau y l a thc h s i xng :

9z6 18z5 73z4 + 164z3 73z2 18z + 9,z8 + 4z6 10z4 + 4z2 + 1,10z6 + z5 47z4 47z3 + z2 + 10z.

nh l 2.1 (Theo [2]). a thc f(z) bc n l a thc i xng khi v ch

khi

znf(1

z) = f(z), z 6= 0. (2.2)

Chng minh. Gi s f(z) c dng

f(z) = a0zn + a1z

n1 + ...+ an. (2.3)

Vi z 6= 0, trong (2.3) thay z bi 1z, ta c

znf(1

z) = anz

n + an1zn1 + ...+ a1z + a0. (2.4)

So snh (2.3) v (2.4) ta thy h thc (2.2) xy ra khi v ch khi

a0 = an, a1 = an1, a2 = an2, ...

ngha l f(z) l a thc i xng. nh l c chng minh.

nh ngha 2.2 (Theo [2]). Cc a thc

a0z2n +a1z

2n1 + ...+an1zn+1 +anzn +an1zn1 + ...+n1a1z+na0,a0z

2n+1 + a1z2n + ...+ an1zn+2 + anzn+1 + anzn + 2an1zn1 + ...+

2n1a1z + 2n+1a0,


trong a0 6= 0 v 6= 0 c gi l cc a thc hi quy. Phng trnhca a thc hi quy c gi l phng trnh hi quy.

Khi = 1 th a thc hi quy tr thnh a thc h s i xng. V d,

phng trnh

2x5 + 6x4 2x3 + 4x2 48x 64 = 0l phng trnh hi quy = 2, cn phng trnh

4x6 + 5x5 3x4 + 10x3 9x2 + 45x+ 108 = 0l phng trnh hi quy = 3.

nh l 2.2 (Theo [2]). Mi a thc hi quy bc chn 2k

f(z) =

a0z2k + a1z

2k1 + ...+ ak1zk+1 + akzk +ak1zk1 + ...+k1a1z+ka0,

u biu din c dng f(z) = zkh(), trong = z +

z, h() l

mt a thc no theo bin v c bc k.

Mi a thc hi quy bc l f(z) u c dng f(z) = (z + )g(z), trong

g(z) l a thc hi quy bc chn.

Chng minh. Trc ht xt a thc h s i xng f(z) c bc 2k. Vi

z 6= 0 ta bin i f(z) nh sau:

f(z) = zk[a0(zk +

k

zk) + a1(z

k1 +k1

zk1) + ...+ ak1(z +

z) + ak].

t = z +

z, sk = z

k +k

zk. Ta s chng t rng sk l a thc bc k

theo . Tht vy, nu t x = z, y =

zth ta c = x + y = 1, =

xy = 2, sk = xk + yk. Do theo nh l 1.1, cc tng ly tha sk l cc

a thc bc k theo cc bin 1, 2, hay l theo cc bin v , ngha l

ch theo bin .

Li xt a thc i xng bc l 2k+1:

f(z) = a0z2k+1 + a1z

2k + ...+ ak1zk+2 + akzk+1 + akzk + 2ak1zk1 +...+ 2k1a1z + 2k+1a0,

Vi z 6= 0 ta bin i f(z) nh sau:f(z) = a0(z

2k+1 + 2k+1) + a1z(z2k1 + 2k1) + ...+ akzk(z + k).


S dng hng ng thc

z2m+1 + 2m+1 = (z + )(z2m z2m1+ ...+ z22m2 z2m1 + 2m),ta c

a0(z2k+1 +2k+1) = a0(z+)(z

2k z2m1+ ...+ z22k2 z2k1 +2k),a1z(z

2k1 + 2k1) = a1(z + )(z2k1 z2m2+ ... z22k3 + z2k2,..............................................................

akzk(z + ) = ak(z + )z

k.

Cng tng v cc ng thc trn v a ra ngoi du ngoc nhn t chung

z + , ta c

f(z) = (z + )g(z),

trong g(z) l tng ca cc a thc

a0(z2k z2m1+ ...+ z22k2 z2k1 + 2k),

a1(z2k1 z2m2+ ... z22k3 + z2k2,

..................................................

akzk.

D dng thy g(z) l a thc hi quy bc 2k. nh l c chng minh.

Bi 2.14 (Theo [5]). Gii phng trnh

9z6 18z5 73z4 + 164z3 73z2 18z + 9 = 0.Li gii. Phng trnh cho l phng trnh i xng bc 6. V z = 0

khng phi l nghim ca phng trnh nn chia hai v ca phng trnh

cho z3 v bin i phng trnh ny v dng

9(z3 +1

z3) 18(z2 + 1

z2) 73(z + 1

z) + 164 = 0

S dng cc cng thc

z +1

z= , z2 +

1

z2= 2 2, z3 + 1

z3= 3 3

ta a c phng trnh trn v dng

93 182 100 + 200 = 0.Nghim ca phng trnh ny l

= 2, = 103, =

10

3.


Do , tm nghim ca phng trnh cho, ta c cc phng trnh

z +1

z= 2, z +

1

z= 10

3, z +

1

z=

10

3.

T cc phng trnh trn ta tm c cc gi tr ca x l nghim ca

phng trnh l:

z = 1, z = 13, z = 3.


2x11 + 7x10 + 15x9 + 14x8 16x7 22x6 22x5 16x4 + 14x3++15x2 + 7x+ 2 = 0.

Li gii. y l phng trnh i xng bc l. Theo nh l 2.2, phng

trnh cho tng ng vi phng trnh

(x+1)(2x10+5x9+10x8+4x720x62x520x4+4x3+10x2+5x+2) = 0Nh vy, phng trnh cho c phn r thnh hai phng trnh

x+ 1 = 0

2x10 + 5x9 + 10x8 + 4x7 20x6 2x5 20x4 + 4x3 + 10x2 + 5x+ 2 = 0Phng trnh th nht c nghim x=-1.

Phng trnh th hai l phng trnh i xng bc 10. V x = 0 khng phi

l nghim ca phng trnh nn chia hai v ca phng trnh cho x5 v

bin i phng trnh ny v dng

2(x5 +1

x5) + 5(x4 +

1

x4) + 10(x3 +

1

x3) + 4(x2 +

1

x2) 20(x+ 1

x) 2 = 0

S dng cc cng thc

x+1

x= , x2 +

1

x2= 2 2, x3 + 1

x3= 3 3,x4 + 1

x4= 4 42 + 2,

x5 +1

x5= 5 53 + 5,

ta a c phng trnh trn v dng

25 + 54 162 40 = 0 (2 + 5)(3 8) = 0.Nghim ca phng trnh ny l

= 0, = 52, = 2.


Do , tm nghim ca phng trnh cho, ta c cc phng trnh

x+1

x= 0, x+

1

x= 5

2, x+

1

x= 2.

T cc phng trnh trn ta tm c cc gi tr ca x l nghim ca

phng trnh cho l:

x = 1, x = 12, x = 2.


2x8 9x7 + 20x6 33x5 + 46x4 66x3 + 80x2 72x+ 32 = 0.Li gii. y l phng trnh bc 8 truy hi vi = 2 v c th vit li

phng trnh dng

2x8 9x7 + 20x6 33x5 + 46x4 33.2x3 + 20.22x2 9.23x+ 2.24 = 0.R rng x = 0 khng phi l nghim ca phng trnh cho. Chia hai v

ca phng trnh cho x4 v bin i v dng

2(x4 +16

x4) 9(x3 + 8

x3) + 20(x2 +

4

x2) 33(x+ 2

x) + 46 = 0.

t = x+

x= x+

2

x. Khi

x2 +4

x2= 2 4, x3 + 8

x3= 3 6, x4 + 16

x4= 4 82 + 8.

Nn phng trnh cui c dng

24 93 + 42 + 21 18 = 0.Phng trnh ny c cc nghim l

= 1, = 2, = 3, = 32, .

Nh vy phng trnh cho tng ng vi t hp cc phng trnh:

x+2

x= 1, x+

2

x= 2, x+

2

x= 3, x+

2

x= 3

2, .

Gii cc phng trnh trn ta tm c nghim ca phng trnh cho l

x = 1, x = 2.


Bi 2.17 (IMO, 1982, Hungari ngh (Theo [2])). Hy xc nh tt c

cc tham s a sao cho phng trnh

16x4 ax3 + (2a+ 17)x2 ax+ 16 = 0.c bn nghim thc lp thnh mt cp s nhn.

Li gii. D thy rng x = 0 khng phi l nghim ca phng trnh

cho. Vi x 6= 0, phng trnh cho tng ng vi

16(x2 +1

x2) a(x+ 1

x) + 2a+ 17 = 0.

t t = x+1

x. Khi |t| 2 v phng trnh trn tr thnh

16t2 at+ 2a 15 = 0.Trc ht ta tm iu kin cn ca tham s a. Gi s phng trnh

cho c 4 nghim thc lp thnh mt cp s nhn. Khi phng trnh

cui cng phi c hai nghim t1, t2, trong t1 cho hai nghim x1,1

x1,

cn nghim t2 cho hai nghim x2,1

x2. Khng mt tnh tng qut, ta gi s

|x1| 1, |x2| 1. Khi c cp s nhn x1, x2, 1x2,

1

x1. Theo tnh cht

ca cp s nhn ta c

x1x2

= x22 x1 = x32 x1 +1

x1= x32 +

1

x32 t1 = t32 3t2.

Mt khc, theo nh l Vite ta c

t1 + t2 =a

16, t1t2 =

2a 1516

.

T ta tm c a=170. Vi a=170 phng trnh cho tr thnh

16x4 170x3 + 357x2 170x+ 16 = 0.Phng trnh ny c bn nghim thc lp thnh mt cp s nhn l :

8, 2,1

2,1

8.

Vy gi tr ca tham s a l a=170.


2.4 Gii h phng trnh

2.4.1 H phng trnh i xng hai n v ng dng

Bi 2.18 (Theo [5]). Gii h phng trnh x5 + y5

x3 + y3=

31

7x2 + xy + y2 = 3

Li gii. Vi iu kin x, y 6= 0. t x+ y = 1, xy = 2. Ta cx2 + y2 = s2 =

21 22, x3 + y3 = s3 = 31 312,

x5 + y5 = s5 = 51 5312 + 5122

Do ta c h{7(51 5312 + 5122) = 23(31 312),21 2 = 3.

T h phng trnh ny, thc hin php th v gii phng trnh ta tm

c

{1 = 12 = 2 hoc

1 = 6

7

2 =15

7

Khi x, y l cc nghim ca cc h phng trnh

{x+ y = 1xy = 2 hoc

x+ y = 6

7

xy =15

7

Gii cc h phng trnh ny ta c cc nghim ca h cho l{x1 = 2,y1 = 1, ;

{x2 = 1,y2 = 2,

{x3 = 2,y3 = 1,

;

{x4 = 1,y4 = 2.

Bi 2.19 (Theo [5]). Gii h phng trnh{x+ y = ax4 + y4 = a4

Li gii. t x+ y = 1, xy = 2.

Ta c


x4 + y4 = s4 = 41 4212 + 222

Do ta c h {1 = a,41 4212 + 222 = a4.

T h phng trnh ny, thc hin php th v gii phng trnh ta tm

c {1 = a2 = 0;

hoc

{1 = a2 = 2a

2.

Khi x, y l cc nghim ca cc h phng trnh{x+ y = axy = 0 hoc

{x+ y = axy = 2a2

Gii cc h phng trnh ny ta c cc nghim ca h cho l{x1 = a,y1 = 0,

;

{x2 = 0,y2 = a.

Bi 2.20 (Theo [5]). Gii h phng trnhx

a+y

b= 1

a

x+b

y= 4

Li gii. Phng trnh cho khng phi l l h i xng, tuy nhin bng

cch tx

a= u,

y

b= v ta c h i xng{

u+ v = 11

u+

1

v= 4

t u+ v = 1, uv = 2, ta c h phng trnh{1 = 112

= 4 {1 = 1

2 =1

4

T ta c h phng trnh{u+ v = 1

u.v =1

4

u =

1

2

v =1

2

x =

a

2

y =b

2


Bi 2.21 (Theo [4]). Gii h phng trnh{x5 y5 = 3093x y = 3

Li gii. Phng trnh cho khng phi l l h i xng, tuy nhin bng

cch t y = z ta c h i xng{x5 + z5 = 3093x+ z = 3

t x+ z = 1, xz = 2, ta c h phng trnh{51 5312 + 5122 = 30931 = 3

{1 = 32 = 10 hoc

{1 = 32 = 19

T ta c x, y l nghim ca cc h phng trnh{x+ z = 3xz = 10 hoc

{x+ z = 3xz = 19

Gii cc h phng trnh trn ta c{x1 = 5z1 = 2 ;

{x2 = 2z2 = 5.

T ta c nghim ca h phng trnh cho l{x1 = 5y1 = 2

;

{x2 = 2y2 = 5.

Bi 2.22 (Theo [3]). Gii h phng trnh{4y3 1 +x = 3

x2 + y3 = 82.

Li gii. tx = u, 4

y3 1 = v ta c h phng trnh{

u+ v = 3u4 + (v4 + 1) = 82

Gii h phng trnh trn, ta c{u1 = 3v1 = 0

;

{u2 = 0v2 = 3.

T ta c nghim ca h phng trnh cho l


{x1 = 9y1 = 1

;

{x2 = 0y2 =

3

83.


5

1

2+ x+ 5

1

2 x = 1

Li gii. t 5

1

2+ x = u, 5

1

2 x = v. Khi ta c h{u+ v = 1u5 + v5 = 1

Gii h phng trnh trn, tm c

{u1 = 1,v1 = 0;

{u2 = 0v2 = 1.

T kt qu trn v u5 =1

2+ x tm c cc nghim ca phng trnh

cho l x1 =1

2, x2 = 1

2.

Vy nghim ca phng trnh cho l x1 =1

2, x2 = 1

2.


x+ 2

17 x2 + x 217 x2 = 9Li gii. t 2

17 x2 = y. Khi ta c h{

x2 + y2 = 17x+ y + xy = 9

Gii h phng trnh trn,tm c

{x1 = 1,y1 = 4;

{x2 = 4y2 = 1.

T kt qu trn ta tm c cc nghim ca phng trnh cho l

x1 = 1, x2 = 4.

Vy nghim ca phng trnh cho l x1 = 1, x2 = 4.


x.19 xx+ 1

.(x+19 xx+ 1

) = 84.

Li gii. t19 xx+ 1

= y th 19 x = xy + y. Khi ta c h{x+ y + xy = 19xy(x+ y) = 84


Gii h phng trnh trn,tm c{x1 = 3,y1 = 4;

{x2 = 4y2 = 3.

{x3 = 6 +

29,

y3 = 6

29;

{x4 = 6

29

y4 = 6 +

29.


x1 = 3, x2 = 4, x3 = 6 +

29, x4 = 6

29.

Vy nghim ca phng trnh cho l

x1 = 3, x2 = 4, x3 = 6 +

29, x4 = 6

29.


x+xx2 1 =

35

12.

Li gii. Vi iu kin |x| > 1,t 1x

= u,

x2 1x

= v . Khi ta c h{u2 + v2 = 11

u+

1

v=

35

12

Gii h phng trnh trn,tm cu1 =

4

5,

y1 =3

5;

u2 =

3

5

v2 =4

5.


x1 =3

5, x2 =

4

5.

Vy nghim ca phng trnh cho l x1 =3

5, x2 =

4

5.

2.4.2 H phng trnh i xng ba n

Gi s P(x,y,z), Q(x,y,z), R(x,y,z) l cc a thc i xng. Xt h

phng trnh {P (x, y, z) = 0Q(x, y, z) = 0R(x, y, z) = 0

(2.5)

Bng cch t

x+ y + z = 1, xy + yz + zx = 2, xyz = 3,.


trn c s cc nh l 1.9, 1.10 ta a h 2.5 v dng{p(1, 2, 3) = 0q(1, 2, 3) = 0r(1, 2, 3) = 0

(2.6)

H phng trnh (2.6) thng n gin hn h (2.5) v c th d dng

tm c nghim 1, 2, 3. Sau khi tm c cc gi tr ca 1, 2, 3, cn

phi tm cc gi tr ca cc n s x, y, z. iu ny d dng thc hin c

nh nh l sau y

nh l 2.3 (Theo [2]). Gi s 1, 2, 3 l cc s thc no . Khi

phng trnh bc ba

u3 1u2 + 2u 3 = 0 (2.7)v h phng trnh {

x+ y + z = 1,xy + xz + yz = 2,xyz = 3.

(2.8)

lin h vi nhau nh sau: nu u1, u2, u3 l cc nghim ca phng trnh

(2.7), th h (2.8) c cc nghim{x1 = u1,y1 = u2,z1 = u3;

{x2 = u1,y2 = u3,z2 = u2;

{x3 = u2,y3 = u1,z3 = u3;{

x4 = u2,y4 = u3,z4 = u1;

{x5 = u3,y5 = u1,z5 = u2;

{x6 = u3,y6 = u2,z6 = u1.

v ngoi ra khng cn cc nghim no khc. Ngc li, nu x=a, y=b, z=c

l nghim ca h (2.8) th cc s a, b, c l nghim ca phng trnh (2.7).

Chng minh. Gi s u1, u2, u3 l cc nghim ca phng trnh (2.7). Khi

ta c ng nht thc

u3 1u2 + 2u 3 = (u u1)(u u2)(u u3).T ta c cc h thc Vite:{

u1 + u2 + u3 = 1,u1u2 + u1u3 + u2u3 = 2,u1u2u3 = 3.

Suy ra u1, u2, u3 l nghim ca h (2.8). Ngoi ra cn nm nghim na

nhn c bng cch hon v cc gi tr ca cc n s. Vn h (2.8)

khng cn nghim no khc c chng t nh sau

Gi s x=a, y=b, z=c l nghim ca h (2.8), ngha l


{a+ b+ c = 1,ab+ bc+ ca = 2,abc = 3.

Khi ta c

u3 1u2 + 2u 3 = u3 (a+ b+ c)u2 + (ab+ bc+ ca)u abc =(u a)(u b)(u c).

iu chng t rng cc s a, b, c l nghim ca phng trnh bc ba

(2.7). nh l c chng minh.

nh l 2.4 (Theo [2]). Gi s 1, 2, 3 l cc s thc cho. cc s

x, y, z xc nh bi h phng trnh (2.8) l cc s thc, iu kin cn v

l

4 = 4313 + 2122 + 18123 432 273 0. (2.9)Ngoi ra, cc s x, y, z l khng m th

1 0, 2 0, 3 0.Chng minh. Gi s x, y, z l nghim ca h (2.8). Khi theo nh l

(2.3) x, y, z l cc nghim ca phng trnh (2.7). Phng trnh (2.7) c

nghim thc khi v ch khi bit thc ca n khng m, ngha l (2.9) c

tha mn. Ngoi ra, nu cc s x, y, z khng m, th hin nhin i 0(i=1,2,3). Ngc li, nu i 0 (i=1,2,3) v (2.9) c tha mn, thphng trnh (2.7) khng th c nghim m. Tht vy, trong (2.7) thay

u=-v ta c phng trnh

v3 + 1v2 + 2v + 3 = 0 (2.10)

V i 0 (i=1,2,3), nn phng trnh (2.10) khng th c nghim dng,do phng trnh (2.7) khng th c nghim m. T suy ra x, y, z l

cc s khng m. nh l c chng minh.

Bi 2.27 (Theo [5]). Gii h phng trnhx3 + y3 + z3 =

73

8,

xy + yz + zx = x+ y + z,xyz = 1.

Li gii. t x + y + z = 1, xy + yz + zx = 2, xyz = 3. H phng

trnh cho tr thnh


s3 =

31 312 + 33 =

73

8,

2 = 1,3 = 1.

Gii h phng trnh ny ta tm c 1 = 2 =7

2, 3 = 1. Theo nh l

(2.3), ta c x, y, z l nghim ca phng trnh

u3 72u2 +

7

2u 1 = 0 (u 1)(u2 5

2u+ 1) = 0

Nghim ca phng trnh ny l u1 = 1, u2 = 2, u3 =1

2. T suy ra

nghim ca h cho l cc b (x, y, z):

(1, 2,1

2), (2, 1,

1

2), (2,

1

2, 1), (1,

1

2, 2), (

1

2, 1, 2), (

1

2, 2, 1).

Bi 2.28 (Theo [5]). Gii h phng trnhx+ y + z =

13

3,

1

x+

1

y+

1

z=

13

3,

xyz = 1.


trnh cho tr thnh 1 =

13

3,

23

=13

3,

3 = 1.

Gii h phng trnh ny ta tm c 1 = 2 =13

3, 3 = 1. Theo nh l

(2.3), ta c x, y, z l nghim ca phng trnh

u3 133u2 +

13

3u 1 = 0 (u 1)(u2 10

3u+ 1) = 0

Nghim ca phng trnh ny l u1 = 1, u2 = 3, u3 =1

3. T suy ra

nghim ca h cho l cc b (x, y, z):

(1, 3,1

3), (3, 1,

1

3), (3,

1

3, 1), (1,

1

3, 3), (

1

3, 1, 3), (

1

3, 3, 1).


Bi 2.29 (Theo [2]). Gii h phng trnh{x+ y + z = a,x2 + y2 + z2 = b2,x3 + y3 + z3 = a3.

trong a, b l cc s thc cho trc.


trnh cho tr thnh {1 = a,21 22 = b2,31 312 + 33 = a3.

Gii h phng trnh ny ta tm c1 = a,

2 =1

2(a2 b2),

3 =1

2a(a2 b2).

Theo nh l (2.3), ta c x, y, z l nghim ca phng trnh

u3 au2 + 12

(a2 b2)u 12a(a2 b2) = 0

(u a)[u2 + 12

(a2 b2)] = 0

T phng trnh trn, ta c

a) Nu |a| > |b|, th phng trnh trn ch c mt nghim thc u=a, do h cho khng c nghim thc. Trong phm vi s phc, th phng trnh c

cc nghim u1 = a, u2 = i

a2 b2

2, u3 = i

a2 b2

2, trong i l n v

o. Khi h cho c nghim (x,y,z) l b s (a, i

a2 b2

2,i

a2 b2

2)

v tt c cc hon v ca n.

b) Nu |a| |b| , th phng trnh trn c ba nghim thc u1 = a, u2 =b2 a2

2, u3 =

b2 a2

2. Khi h phng trnh cho c nghim

(x,y,z) l b s (a,

b2 a2

2,

b2 a2

2)v tt c cc hon v ca n.


2.5 Tm nghim nguyn ca cc phng trnh i

xng

Bi 2.30 (Theo [2]). Tm nghim nguyn ca phng trnh

x3 + y3 + 1 = 3xy

Li gii. t 1 = x+ y, 2 = xy. Phng trnh tr thnh

31 312 + 1 = 32 (1 + 1)(21 1 + 1 32) = 0.Trng hp 1: 1 + 1 = 0, ta c x + y + 1 = 0, phng trnh c v s

nghim nguyn (x Z, y = 1 x).Trng hp 2: 21 1 + 1 32 = 0. Ta vit phng trnh ny di dng

21 1 + 1 = 32T vic t 1 = x + y, 2 = xy, th iu kin tn ti hai s x, y l

1 42. S dng iu kin ny ta c

21 1 + 1 3

421 21 41 + 4 0

(1 2)2 0 1 = 2 2 = 1.

Trong trng hp ny ta c h{x+ y = 2,xy = 1.

H ny lun c nghim nguyn duy nht l x = y = 1. Nh vy, nghim

ca phng trnh cho l{x = 1,y = 1;

{x Z,y = 1 x.

Bi 2.31. Tm nghim nguyn ca phng trnh

x+ y = x2 xy + y2

Li gii. t 1 = x+ y, 2 = xy. Phng trnh tr thnh

1 = s2 22 1 = 21 32 21 1 = 32.T vic t 1 = x + y, 2 = xy, th iu kin tn ti hai s x, y l

1 42. S dng iu kin ny ta c42S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

21 1 3

421 21 41 0 0 1 4

Mt khc, t phng trnh

21 1 = 32. (21 1)2 = 122 + 1

thy 122 + 1 l s chnh phng, nn t chn c

+) 1 = 2 = 0, phng trnh c nghim nguyn: x = y = 0.

+) 1 = 1, 2 = 0, phng trnh c cc nghim nguyn l{x = 1,y = 0;

{x = 0,y = 1.

+) 1 = 2, 122 + 1 = 9, phng trnh khng c nghim nguyn.


{x = 2,y = 1.


Vy phng trnh cho c cc nghim nguyn l:{x = 0,y = 0;

{x = 0,y = 1. ;

{x = 1,y = 0. ;

{x = 1,y = 2. ;

{x = 2,y = 1. ;

{x = 2,y = 2.

Bi 2.32 (Theo [1]). Tm cc nghim nguyn dng ca h phng trnh{x+ y = z,x3 + y3 = z2.

Li gii.t 1 = x+ y, 2 = xy. H phng trnh tr thnh{1 = z,s3 = z

2. {1 = z,31 312 = z2.

Th phng trnh th nht vo phng trnh th hai, ta c

31 312 = 21 1(21 1 32) = 0Trng hp 1: 1 = 0 z = 0 x = y, r rng khng tha mn, donghim cn tm nguyn dng.

Trng hp 2:


21 1 32 = 0Tng t bi tp 2.30, d dng tm c cc nghim nguyn dng ca h

l {x = 2,y = 1,z = 3;

{x = 1,y = 2,z = 3

;

{x = 2,y = 2z = 2.

2.6 Chng minh cc ng thc

Bi 2.33. Cho x+ y = 1, x3 + y3 = a, x5 + y5 = b. Chng minh rng

5a(a+ 1) = 9b+ 1

Li gii. t 1 = x+ y = 1, 2 = xy. Ta c:

x3 + y3 = a 31 312 = a 2 =1 a

3

V

x5 + y5 = s5 = 51 5312 + 5122

nn

b = 1 5.1 a3

+ 5.(1 a)2

9

hay

5a(a+ 1) = 9b+ 1

Bi 2.34 (Theo [5]). Chng minh ng nht thc

(x+ y)3 + 3xy(1 x y) 1 = (x+ y 1)(x2 + y2 xy + x+ y + 1).Li gii. Theo cng thc Waring, ta c

(x+ y)3 + 3xy(1 x y) 1 = 31 + 32(1 1) 1 == 31 + 32 312 1.

Mt khc, cng c

(x+ y 1)(x2 + y2 xy + x+ y + 1) = (1 1)(21 32 + 1 + 1) == 31 312 + 21 + 1 21 + 32 1 1 =

= 31 + 32 312 1.T hai h thc trn, ta c iu phi chng minh.



(x+ y)7 x7 y7 = 7xy(x+ y)(x2 + xy + y2)2.Li gii. Theo cng thc Waring, ta c

(x+ y)7 x7 y7 = 71 s7 = 71 (71 7512 + 143122 7132) == 7512 143122 + 7132 = 712(41 2212 + 22) =

= 712(21 2)2 = 7xy(x+ y)(x2 + xy + y2)2.


(x+ y + z)(xy + yz + zx) xyz = (x+ y)(y + z)(z + x).Li gii. S dng cc cng thc qu o, khai trin v phi, ta c

(x+ y)(y + z)(z + x) = x2y + x2z + y2x+ y2z + z2x+ z2y + 2xyz =

= O(x2y) + 23 = (12 33) + 23 = 12 3 == (x+ y + z)(xy + yz + zx) xyz.

Bi 2.37 (Theo [5]). Chng minh rng, nu

x+ y + z = xy + yz + zx = 0.

th

3(x3y3 + y3z3 + z3x3) = (x3 + y3 + z3)2.

Li gii. S dng cc cng thc qu o, khai trin v phi vi lu iu

kin ca bi l 1 = 0, 2 = 0, ta c

3(x3y3 + y3z3 + z3x3) = 3O(x3y3) = 3(32 + 323 3123) = 923.

Tng t vi v tri, ta c

(x3 + y3 + z3)2 = s23 = (31 312 + 33)2 = 923.

T cc khai trin trn, suy ra iu phi chng minh.

Bi 2.38 (Theo [2]). Chng minh rng, nu cc s thc x, y, z, a, b, c

tha mn cc h thc{x+ y + z = a+ b+ c,x2 + y2 + z2 = a2 + b2 + c2,x3 + y3 + z3 = a3 + b3 + c3.

th vi mi s t nhin n:


xn + yn + zn = an + bn + cn.

Li gii. K hiu 1, 2, 3 l cc a thc i xng c s theo cc bin x,

y, z; cn 1, 2, 3 l cc a thc i xng c s theo cc bin a, b, c. S

dng cc cng thc Waring, theo gi thit ta c

1 = 1,

21 22 = 21 22,31 312 + 33 = 31 312 + 33,

Suy ra

1 = 1,2 = 2,3 = 3,

Khi vi mi a thc (t1, t2, t3) ta c

(t1, t2, t3) = (1, 2, 3).

Gi s f(x,y,z) l mt a thc i xng v theo nh l duy nht f(x, y, z) =

(1, 2, 3), f(a, b, c) = (1, 2, 3). T suy ra f(x, y, z) = f(a, b, c).

Trong trng hp ring ta c

xn + yn + zn = an + bn + cn.

Bi 2.39 (Theo [5]). Chng minh rng nu a + b + c = 0, th cc ng

sau y ng

1. a3 + b3 + c3 = 3abc;

2. a3 + b3 + c3 + 3(a+ b)(b+ c)(c+ a) = 0;

3. a2(b+c)2+b2(c+a)2+c2(a+b)2+(a2+b2+c2)(ab+bc+ca) = 0;

4. a4 + b4 + c4 = 2(a2b2 + b2c2 + c2a2) =1

2(a2 + b2 + c2)2;

5.a5 + b5 + c5

5=a3 + b3 + c3

3.a2 + b2 + c2

2= abc.

a2 + b2 + c2

2;

6.a7 + b7 + c7

7=a5 + b5 + c5

5.a2 + b2 + c2

2=a4 + b4 + c4

2.a3 + b3 + c3

3;

7.a7 + b7 + c7

7.a3 + b3 + c3

3= (

a5 + b5 + c5

5)2,

8. (a7 + b7 + c7

7)2 = (

a5 + b5 + c5

5)2.a4 + b4 + c4

2Li gii. K hiu 1, 2, 3 v sk tng ng l cc a thc i xng c s

v tng ly tha ca cc bin a, b, c. Theo cng thc Waring ta c


Cng thc tnh sk = xk + yk + zk, theo 2 3 khi c 1 = 0

a+ b+ c = s1 = 1 = 0;a2 + b2 + c2 = s2 = 22;a3 + b3 + c3 = s3 = 33;a4 + b4 + c4 = s4 = 2

22;

a5 + b5 + c5 = s5 = 523;a6 + b6 + c6 = s6 = 3

23 232;

a7 + b7 + c7 = s7 = 7223;

a8 + b8 + c8 = s8 = 242 8223;

a9 + b9 + c9 = s9 = 333 9323;

a10 + b10 + c10 = s10 = 252 + 152223;.........................................................

Li gii.

1. Theo cng thc trn, hin nhin ng.

2. Theo cng thc trn v kt qu bi 2.28, hin nhin ng.

3. a2(b+ c)2 + b2(c+ a)2 + c2(a+ b)2 + (a2 + b2 + c2)(ab+ bc+ ca) =

= 2O(a2b2) + 2O(a2bc) + s22 = 222 2.0 + (22)2 = 0;

4. T cng thc trn, ta c

a4 + b4 + c4 = s4 = 222

2(a2b2 + b2c2 + c2a2) = 2.O(a2b2) = 2(22 213) = 2221

2(a2 + b2 + c2)2 =

1

2s22 =

1

2(22)2 = 222

T suy ra iu phi chng minh.

5.a3 + b3 + c3

3.a2 + b2 + c2

2= 2.3 = a

5 + b5 + c5

5=

= 3.22

2= abc.

a2 + b2 + c2

2;

6.a5 + b5 + c5

5.a2 + b2 + c2

2= (23).(2) = 223 =

a7 + b7 + c7

7=

=2222.3 =

a4 + b4 + c4

2.a3 + b3 + c3

3;

7.a7 + b7 + c7

7.a3 + b3 + c3

3= 223.3 = (23)

2 = (a5 + b5 + c5

5)2,

8.(a7 + b7 + c7

7)2 = (223)

2 = (23)2.22 = (

a5 + b5 + c5

5)2.a4 + b4 + c4

2.


(a+ b)7 a7 b7(a+ b)3 a3 b3 =

7

6[(a+ b)4 + a4 + b4].


Li gii. t x = a, y = b, z = a b. Khi 1 = a+ b+ (a b) = 0,nn theo cng thc Waring ta c

(a+ b)7 a7 b7(a+ b)3 a3 b3 =

z7 x7 y7z3 x3 y3 =

s7s3

=722333

=7

322.

Ta bin i v phi ca ng thc cn chng minh nh sau:

7

6[(a+ b)4 + a4 + b4] =

7

6[(z)4 + a4 + b4] = 7

6s4 =

7

6.222 =

7

3.22.


Phng php trnh by trn thng c p dng khi trong bi ton

c cc hiu ab, bc, ca. Khi nu t x = ab, y = bc, z = cath 1 = x + y + z = 0. Trong trng hp ny cc cng thc Waring i

vi tng ly tha tr nn n gin hn rt nhiu. Xt cc v d sau.


(b c)3 + (c a)3 + (a b)3 3(b c)(c a)(a b) = 0.Li gii.

t x = a b, y = b c, z = c a th 1 = x + y + z = 0 v ng thccn chng minh tr thnh

x3 + y3 + z3 3xyz = s3 33 = 33 33 = 0.Bi 2.42 (Theo [5]). Chng minh ng nht thc

25[(b c)7 + (c a)7 + (a b)7].[(b c)3 + (c a)3 + (a b)3] =21[(b c)5 + (c a)5 + (a b)5]2.

Li gii. t x = a b, y = b c, z = c a th 1 = x + y + z = 0 vng thc cn chng minh tr thnh

25(x7 + y7 + z7).(x3 + y3 + z3) = 21(x5 + y5 + z5)2

hay

25s7.s3 = 21s25

S dng cc cng thc ca tng ly tha vi ch 1 = 0, ta c

25s7.s3 = 25.7223.33 = 525(23)

2; 21s25 = 21(523)2 = 525(23)2



Bi 2.43 (Theo [5]). Vi a+ b+ c = 2p, chng minh rng

1. a(p b)(p c) + b(p a)(p c) + c(p a)(p b)++2(p a)(p b)(p c) = abc.

2. (p a)3 + (p b)3 + (p c)3 + 3abc = p3Li gii

1. Khai trin v phi, s dng cng thc Waring vi ch

1 = a+ b+ c = 2p, ta c

p2(a+ b+ c) 2p(ab+ ac+ bc) + 3abc++2[p3 p2(a+ b+ c) + p(ab+ ac+ bc) abc] =

= (12

)21 12 + 33 + 2(31

8+122 3) = 3 = abc.

2. Vit ng thc cn chng minh nh sau

(a+ b+ c

2)3 + (

a b+ c2

)3 + (a+ b c

2)3 + 3abc = (

a+ b+ c

2)3.

Theo kt qu ca bi 2.12, suy ra iu phi chng minh.

Bi 2.44 (Theo [5]). Chng minh rng, nu xy + yz + zx = 1, th

x

1 x2 +y

1 y2 +z

1 z2 =4xyz

(1 x2)(1 y2)(1 z2) .

Li gii. Bin i biu thc cho, ta nhn c biu thc sau

x(1 y2)(1 z2) + y(1 z2)(1 x2) + z(1 x2)(1 y2) = 4xyz.S dng cc cng thc ca tng ly tha v qu o vi ch

2 = xy + yz + zx = 1 bin i v phi ng thc trn ta c

O(x)O(x2y) +O(x2y2z) = 1 (1 33) + 3 = 4xyz.Bi 2.45 (Theo [2]). Chng minh rng, nu

1

a+

1

b+

1

c=

1

a+ b+ c,

th vi mi s l n, ta c

(1

a+

1

b+

1

c)n =

1

an + bn + cn=

1

(a+ b+ c)n.

Li gii. T gi thit ca bi ton, ta c

23

=1

1,


hay

12 3 = 0,Theo kt qu ca bi 2.11 ta c ng thc

(a+ b)(a+ c)(b+ c) = 0,

T ng thc trn, ta c

a = b, a = c, b = c.Thay vo ng thc cn chng minh, ta thy ng thc

(1

a+

1

b+

1

c)n =

1

an + bn + cn=

1

(a+ b+ c)n.

lun ng vi mi s l n.

2.7 Chng minh bt ng thc

Vi hai s thc x, y, t x+ y = 1, xy = 2. Khi ta c

(x+ y)2 0 (x+ y)2 4xy 0 21 42 0.Mnh 2.1 (Theo [2]). Cho x, y R. t x+ y = 1, xy = 2. Khi

1 42. (2.11)ng thc xy ra khi v ch khi x = y.

Gi s cn chng minh bt ng thc f(x, y) 0 (vi mi x 0, y 0hoc x + y = a, ty thuc vo bi ton), trong f(x, y) l a thc i

xng. Trc ht ta biu din f(x, y) theo 1, 2. Sau trong a thc va

nhn c thay 2 bi 1 bng cch t 2 =1

4(21 z) vi z 0. Hoc

cng c th biu din 1 bi 2 bng cch vit 21 = z + 42.

Bi 2.46 (Theo [2]). Chng minh rng, nu a,b v c l ba s thc tha

mn iu kin a+ b c 0, th

a2 + b2 c2

2, a4 + b4 c

4

8, a8 + b8 c

8

128

Li gii. t a+ b = 1, ab = 2.Ta c


s2 = a2 + b2 = 21 22 = 21 2.

1

4(21 z) =

1

221 +

1

2z,

v z 0 v theo gi thit 1 c 0, nn s2 12c2, ngha l

a2 + b2 c2

2

Vn dng bt ng thc trn , ta c

a4 + b4 = (a2)2 + (b2)2 12

(1

2c2)2 c

4

8.

Tng t d dng chng minh

a8 + b8 c8

128

Mt cch tng qut, ta c bi ton

Nu a, b v c l ba s thc tha mn iu kin a+ b c 0, th vi min nguyn dng, ta c

a2n + b2n 122n1

.c2n

Lu : Cho c nhn cc gi tr thc khc nhau, ta nhn c cc bi

ton khc nhau. Chng hn cho c=1, ta c bi ton

Chng minh rng, nu a, b l cc s thc tha mn iu kin a + b 1,th

a2 + b2 12, a4 + b4 1

8, a8 + b8 1

128

Bi 2.47 (Theo [5]). Cho a, b R. Chng minh rng1. 8(a4 + b4) (a+ b)4,2. a6 + b6 a5b+ ab5Li gii

1. Ta c

8(a4 + b4) (a+ b)4 = 8s4 41 = 8(41 4212 + 222) 41 == 741 3221.

1

4(21 z) + 16.

1

16(21 z)2 = 621z + z2 0

2. Ta c


a6 + b6 a5b+ ab5 = s6 2s4 = 61 7412 + 13 1222 432 == 61 741.

1

4(21 z) + 1321.

1

16(21 z)2 4.

1

64(21 z)3 =

=5

1641z +

5

821z

2 +1

16z3 0

Bi 2.48 (Theo [5]). Chng minh rng vi mi a, b khng m c cc bt

ng thc

1.

a2

b+

b2

a a+b,

2. (a+b)8 64ab(a+ b)2.

Li gii.

1. ta = u,

b = v. Ta c bt ng thc

u2

v+v2

u u+ v hay u3 + v3 uv(u+ v)

T gi thit ca bi ton, ta c u 0, v 0 t cng c1 0, z = 1 42 0

Do

u3 +v3uv(u+v) = 3131212 = 31412 = 1(2142) 0.2. t

a = u,

b = v. Ta c bt ng thc (u+ v)8 64u2v2(u2 + v2)2.

T gi thit ca bi ton, ta c u 0, v 0, do bt ng thc trn cth vit thnh

(u+ v)4 8uv(u2 + v2)Ta c:

(u+ v)4 8uv(u2 + v2) = 41 82(21 22) = 41 8212 + 1622 == 41 821.

1

4(21 z) + 16.

1

16(21 z)2 = z2 0.

T ta c iu phi chng minh.

Vi cc b s thc (x, y, z) hay (a, b, c) v.v.. ta lun hiu 1, 2, 3 l cc

a thc i xng c s ca cc b . Chng hn vi b s thc (x, y, z)

ta c

1 = x+ y + z, 2 = xy + xz + yz, 3 = xyz.

Mnh 2.2 (Theo [2]). Vi cc s thc x, y, z lun c :

a) 21 32 b) 22 313. (2.12)Du ng thc xy ra khi v ch khi x=y=z.


Chng minh. Vi mi s thc x, y, z ta lun c bt ng thc

(x y)2 + (y z)2 + (z x)2 0.Du ng thc xy ra khi v ch khi x=y=z. Khai trin v tri ca bt

ng trn v thu gn ta c s2 2 0. Thay s2 = 21 22, ta c btng thc th nht trong (2.12).

Theo bt ng thc th nht trong (2.12), ta c

(x+ y + z)2 3(xy + xz + yz).Trong bt ng trn thay x = ab, y = ac, z = bc ta c

(ab+ ac+ bc)2 3(a2bc+ ab2c+ abc2) = 3abc(a+ b+ c).T suy ra bt ng thc th hai trong (2.12) vi cc s thc a, b, c.

Mnh 2.3 (Theo [2]). Vi cc s thc dng x, y, z lun c :

a) 12 93 b) 31 273 c) 32 2723. (2.13)Du ng thc xy ra khi v ch khi x = y = z.

Chng minh. V x, y, z l cc s thc dng nn 1, 2, 3 cng l cc s

dng. Nhn hai v tng ng ca cc bt ng thc trong (2.12) ta c

2122 9123. Gin c hai v ta nhn c bt ng thc 12 93.

T bt ng thc th nht trong (2.12), ta c

41 = 21.

21 21.32 = 31(12) 31.93 = 2713.

Gin c hai v bt ng thc trn, vi lu 1 dng, ta nhn c bt

ng thc 31 273.T bt ng thc th hai trong (2.12) v bt ng thc th nht trong

(2.13) ta c

32 = 2.22 2.313 = 33(12) 33.93 = 2723.

Nhn xt. Cc bt ng trong (2.13) vn cn ng vi x, y, z l cc s

khng m.

T bt ng thc th nht v th hai trong (2.13), ta c cc bt ng

thc quen thuc

(x+ y + z)(1

x+

1

y+

1

z) 9 v

(x+ y + z)3 27xyz hay x+ y + z3

3xyz.


Mnh 2.4 (Theo [2]). Vi mi s thc khng m x, y, z ta lun c cc

bt ng thc

a) 31 + 93 412 b) 231 + 93 712. (2.14)Du ng thc xy ra khi v ch khi x = y = z.

Chng minh. Trc ht, vi mi x, y, z khng m, ta chng minh c bt

ng thc

x(x y)(x z) + y(y x)(y z) + z(z x)(z y) 0Tht vy, do tnh cht i xng ca x, y, z nn khng lm mt tnh tng

qut ta gi s rng x y z. Khi bt ng thc cho c vit linh sau

(x y)[x(x z) y(y z)] + z(x z)(y z) 0Bt ng thc ny ng.

Khai trin v tri bt ng trn ta c

x(x y)(x z) + y(y x)(y z) + z(z x)(z y) == (x3 + y3 + z3) + 3xyz (x2y + x2z + y2x+ y2z + z2x+ z2y) == s3 + 33 O(x2y) = (31 312 + 33) + 33 (12 33) =

= 31 412 + 93.T ta c kt qu ca bt ng thc th nht trong cng thc (2.14).

T bt ng thc th nht trong (2.12) v (2.14), ta c

231 + 93 = (31 + 93) +

31 412 + 312 = 712.

Vy cc bt ng thc trong (2.14) c chng minh.

Bi 2.49 (Anh, 1999 (Theo [2])). Cho x, y, z l cc s khng m tha mn

iu kin x+ y + z = 1. Chng minh rng

7(xy + yz + zx) 2 + 9xyz.Li gii. V x+y+z = 1, nn ta c th vit bt ng thc cn chng minh

dng

7(x+ y + z)(xy + yz + zx) 2 + 9xyz.t 1 = x+ y+ z, 2 = xy+xz+ yz, 3 = xyz, bt ng thc cn chng

minh tr thnh

712 231 + 9354S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

Theo 2.10 ta c iu phi chng minh.

Bi 2.50 (Theo [5]). Chng minh rng vi mi s thc a, b, c ta c cc bt

ng thc sau

1. a2 + b2 + c2 ab+ bc+ ca,2. a2 + b2 + c2 1

3(a+ b+ c)2,

3. a2b2 + b2c2 + c2a2 abc(a+ b+ c),4. (ab+ bc+ ca)2 3abc(a+ b+ c),5. a2 + b2 + 1 ab+ a+ b,Li gii. t 1 = a+ b+ c, 2 = ab+ bc+ ca, 3 = abc. Ta c

1. Bt ng thc cho tng ng vi 21 22 2 hay 21 32.Theo (2.12) ta c iu phi chng minh.

2. Bt ng thc cho tng ng vi 21 22 1

321 hay

21 32.

Theo (2.12) ta c iu phi chng minh.

3.Bt ng thc cho tng ng vi O(a2b2) 13hay 22 213 13. Theo (2.12) ta c iu phi chng minh.4.Theo (2.12) ta c iu phi chng minh.

5.Trong bt ng thc 1, cho c = 1, ta c iu phi chng minh.

Bi 2.51 (Theo [5]). Chng minh rng vi mi s thc dng a, b, c ta c

cc bt ng thc sau

1. (a+ b+ c)(a2 + b2 + c2) 9abc,2. ab(a+ b 2c) + bc(b+ c 2a) + ac(a+ c 2b) 0,3. ab(a+ b) + ac(a+ c) + bc(b+ c) 6abc,4. 2(a3 + b3 + c3) ab(a+ b) + ac(a+ c) + bc(b+ c),5.

a3 + b3 + c3

a2 + b2 + c2 x+ y + z

3,

Li gii. t 1 = a+ b+ c, 2 = ab+ ac+ bc, 3 = abc.Khi

1. Bt ng thc cho tng ng vi

1(21 22) 93 hay 31 212 + 93.

Theo cng thc (2.12) trong mnh (2.2) v cng thc (2.13) trong mnh

(2.3) th

31 312 v 12 93.T suy ra iu phi chng minh.



ab(1 3c) + bc(1 3a) + ac(1 3b) 0 12 93 0.

Bt ng thc cui cng ng theo cng thc (2.13) trong mnh (2.3).Vy

bt ng thc cho c chng minh.


ab(1 c) + bc(1 a) + ac(1 b) 63 12 33 63 12 93.




2s3 O(a2b) 2(31 312 + 33) 12 33 231 712 + 93 0 231 + 93 712.




3s3 s1s2 3(31 312 + 33) 1(21 22) 231 712 + 93 0 231 + 93 712.



Bi 2.52 (Theo [5]). Chng minh rng, nu a, b, c l di cc cnh ca

mt tam gic th ta c cc bt ng thc

1. 2(ab+ bc+ ca) > a2 + b2 + c2,

2. (a2 + b2 + c2)(a+ b+ c) > 2(a3 + b3 + c3).

Li gii.

1. V a, b, c l ba cnh ca mt tam gic, nn ta c th t

x = a+ b c, y = a b+ c, z = a+ b+ ctrong x, y, z l cc s dng. Khi a, b, c c biu th theo x, y, z nh

sau:

a =x+ y

2, b =

x+ z

2, c =

y + z

2

Thay vo bt ng thc cn chng minh ta c bt ng thc


2(x+ y

2.x+ z

2+x+ y

2.y + z

2+x+ z

2.y + z

2) >

> (x+ y

2)2 + (

x+ z

2)2 + (

y + z

2)2

t 1 = x+ y+ z, 2 = xy+ xz + yz, 3 = xyz, s2 = x2 + y2 + z2 l cc

a thc i xng c s v tng ly tha ca x, y ,z. Ta c

2(s2 + 32) > 2s2 + 22 42 > 0D thy bt ng thc cui cng l bt ng thc ng. Vy bt ng thc

cho c chng minh.

2. V a, b, c l ba cnh ca mt tam gic, nn ta c th t

x = a+ b c, y = a b+ c, z = a+ b+ ctrong x, y, z l cc s dng. Khi a, b, c c biu th theo x, y, z nh

sau:

a =x+ y

2, b =

x+ z

2, c =

y + z

2

Thay vo bt ng thc cn chng minh ta c bt ng thc

[(x+ y

2)2 + (

x+ z

2)2 + (

y + z

2)2](x+ y + z) >

> 2[(x+ y

2)3 + (

x+ z

2)3 + (

y + z

2)3].

t 1 = x+ y + z, 2 = xy + xz + yz, 3 = xyz, s2 = x2 + y2 + z2,

O(x2y) = x2y + x2z + y2x + y2z + z2x + z2y l cc a thc i xng c

s, tng ly tha v cng thc qu o ca x, y, z. Ta c

2(s2 + 22)1 > 2s3 + 3O(x2y)

2(21 22 + 22)1 > 2(31 312 + 33) + 3(12 33) 12 + 3 > 0.

D thy bt ng thc cui cng l bt ng thc ng. Vy bt ng thc

cho c chng minh.


Chng 3

a thc i xng n bin v ngdng

3.1 Cc khi nim

nh ngha 3.1 (Theo [2]). Gi s x = (x1, x2, ..., xn) Rn.a thc f(x) = f(x1, x2, ..., xn) c hiu l hm s c dng

f(x) =mk=0

Mk(x),

trong

Mk(x) = Mk(x1, x2, ..., xn) =

j1+j2+...+jn=k

aj1j2...jnxj11 x

j22 ...x

jnn ,

ji N, (i = 0, 1, 2, ..., n).nh ngha 3.2 (Theo [2]). a thc f(x1, x2, ..., xn) theo cc bin x1, x2, ..., xnc gi l i xng nu n khng thay i khi i ch gia hai bin bt

k.

nh ngha 3.3 (Theo [2]). a thc f(x1, x2, ..., xn) theo cc bin x1, x2, ..., xnc gi l thun nht bc m, nu

f(tx1, tx2, ..., txn) = tmf(x1, x2, ..., xn),t 6= 0.

nh ngha 3.4 (Theo [2]). K hiu

sk = xk1 + x

k2 + ...+ x

kn, k Z.

0 = 1, 1 =ni=1

xi, 2 =n

i,j=1,i

n = x1x2.....xn.

Ta gi sk l cc tng ly tha, cn r(x)(r = 1, 2, 3, ..., n) l cc a thc

i xng s cp bc r.

Chng hn, vi n = 4, ta c

1 = x1 + x2 + x3 + x4,

2 = x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4,

3 = x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4,

4 = x1x2x3x4.

S cc s hng trong a thc i xng bc k: k bng

Ckn =n!

k!(n k)! .

nh ngha 3.5 (Theo [2]). a thc vi s hng ti thiu, mi s hng l

mt n thc c dng xk11 xk22 ...x

knn c gi l qu o ca n thc (orbit)

v c k hiu l

O(xk11 xk22 ...x

knn )

Chng hn vi n = 4, ta c

O(x21x32) = x

21x

32 + x

21x

33 + x

21x

34 + x

22x

31 + x

22x

33 + x

22x

34 + x

23x

31+

+x23x32 + x

23x

34 + x

24x

31 + x

24x

32 + x

24x

33.

T nh ngha v qu o suy ra, nu cc s m k1, k2, ..., kn khc nhau th

s cc s hng ca qu o O(xk11 xk22 ...x

knn ) bng n!. M rng cho trng

hp cc s m k1, k2, ..., kn khng nht thit phi khc nhau ta c khi

nim qu o ton phn.

nh ngha 3.6 (Theo [2]). a thc bng tng tt c cc hon v theo cc

bin x1, x2, ..., xn ca n thc xk11 x

k22 ...x

knn c gi l qu o ton phn

v c k hiu l

O(xk11 x

k22 ...x

knn )

Qu o ton phn ch khc qu o thng thng bi h s, c th l:

Nu trong cc s m k1, k2, ..., kn c m1 s bng nhau, sau li c m2 s

bng nhau,..., v cui cng cn li ml s bng nhau, th

O(xk11 x

k22 ...x

knn ) = m1!m2!....ml!O(x

k11 x

k22 ...x

knn ) (3.1)

T cng thc (3.1) ta thy, nu cc s m k1, k2, ..., kn khc nhau, th qu

o ton phn trng vi qu o thng thng.


Cc cng thc di y cho php biu din cc qu o ton phn qua cc

tng ly tha

O(xk1x

l2) = (n 2)!(sksl sk+l), (3.2)

(n 2)O(xk1xl2xm3 ) = O(xk1xl2)sm O(xk+m1 xl2)O(xk1xl+m2 ), (3.3)......................

(n r)O(xk11 xk22 .....xkrr xkr+1r+1 ) =

= O(xk11 x

k22 .....x

krr )skr+1

O(xk1+kr+11 xk22 .....xkrr )O(xk11 xk2+kr+12 .....xkrr )

..........................

O(xk11 xk22 .....xkr+kr+1r ). (3.4)nh ngha 3.7 (Theo [2]). Xt hai n thc Axk11 x

k22 ...x

knn , Bx

l11 x

l22 ...x

lnn .

Ta ni n thc th nht tri hn n thc th hai, nu k1 > l1; hoc

k1 = l1 v k2 > l2; ...; hoc k1 = l1, k2 = l2, ..., ks = ls v ks+1 > ls+1.

Chng hn, so snh hai a thc f = x31x42x

23, g = x

31x

42x

54 ta vit

chng dng f = x31x42x

23x

04, g = x

31x

42x

03x

54. Theo nh ngha ta thy f tri

hn g.

3.2 Biu din cc tng ly tha qua cc a thc i

xng c s

nh l 3.1 (Cng thc truy hi Newton (Theo [2])). Cc tng ly tha

v cc a thc i xng c s lin h vi nhau theo cng thc

sk = 1sk1 2sk2 + 3sk3 ...+ (1)k1kk (3.5)(trong cng thc (3.5) s hng (1)i1isni = 0 khi i > n).Chng minh. Xt a thc mt bin theo x

P (x) =ni=1

(x xi) = xn 1xn1 + ...+ (1)n1n1x+ (1)nnv cc a thc Qi(x) c xc nh theo cng thc

Qi (x) =n

j=1,j 6=i(x xj) = P (x)

x xi (1 i n)


K hiu

k(x) = k(x1, x2, ..., xn), k(x0i ) = k(x1, x2, ..., xi1, 0, xi+1, ..., xn).

R rng l

k = k(x0i ) + xik1(x

0i ), (3.6)

Qi (x) =n1k=0

(1)kk(x0i )xn1k (3.7)

T (3.6), suy ra

0(x0i ) = 1, 1(x

0i ) = 1 xi,

2(x0i ) = 2(x) xi1(x) + x2i ,

3(x0i ) = 3(x) xi2(x) + x2i1(x) x3i .

Bng phng php quy np ta c cng thc

k(x0i)

=kj=0

(1)jxjikj (x) (3.8)

(k=0, 1, ..., n-1).

Ta c cng thc

ni=1

P (x)

x xi = P (x) =

n1k=0

(1)k (n k)k (x)xnk1. (3.9)

Mt khcni=1

P (x)xxi =

ni=1

Qi (x) =ni=1

n1k=0

(1)kk(x0i)xnk1 =

=n1k=0

(1)k(

ni=1

k(x0i))

xnk1 (3.10)

T (3.9) v (3.10), suy ra

ni=1

k(x0i)

= (n k)k (x) , (0 k n 1) . (3.11)

T (3.9) v (3.11), suy ra

ni=1

kj=0

(1)jxjik(x0j)

= (n k)k (x) , (k n 1) ,


kj=0

(1)jsjkj (x) = (n k)k (x), (k n 1) . (3.12)

Ch rng s0 = n, t (3.12) d dng suy ra (3.5). nh l 3.1 c chng

minh.

Nhn xt. Trn y trnh by mt chng minh ca nh l 3.1 tuy

cha c n gin nhng li cho mt s cng thc cn thit cho sau

ny. nh l trn c th c chng minh bng cch n gin sau. V

x1, x2, ...., xn l nghim ca a thc P(x), nn ta c cc ng thc sau

xn1 1xn11 + 2xn21 ...+ (1)nn = 0xn2 1xn12 + 2xn22 ...+ (1)nn = 0

...............................................

xnn 1xn1n + 2xn2n ...+ (1)nn = 0Cng theo tng ct v vi v cc ng thc trn ta c cng thc (3.5).

S dng cng thc (3.5) d dng thu c cc cng thc sau y

s1 = 1.1;

s2 = 1s1 22;s3 = 1s2 2s1 + 33;s4 = 1s3 2s2 + 3s1 44;s5 = 1s4 2s3 + 3s2 4s1 + 55;s6 = 1s5 2s4 + 3s3 4s2 + 55s1 66;........................................................

T cc cng thc trn, mt cch lin tip ta nhn c cc cng thc

sau

s1 = 1;

s2 = 21 22;

s3 = 31 312 + 33;

s4 = 41 4212 + 222 + 413 44;

s5 = 51 5312 + 5122 + 5213 523 514 + 55;

s6 = 61 6412 + 92122 232 + 6313 12123+

+323 6214 + 624 + 615 66;........................................................



din qua cc a thc i xng c s theo cng thc

skk

=

m1+2m2+...+kmk=k

(1)km1m2...mk.(m1 +m2 + ...+mk 1)!m1!m2!...mk!

.m11 m22 ...

mkk

(3.13)

Cng thc (3.13) c th c chng minh bng phng php quy np vi

s tr giy ca cng thc truy hi (3.5) v c gi l cng thc Waring.

Ngoi ra c th chng minh c rng cc tng ly tha sk c biu

din theo cc a thc i xng c s j bi cng thc

sk = (1)k1

1 0 0 ... 0 111 1 0 ... 0 12 1 1 ... 0 1... ... ... ... ... ...k2 k3 k4 ... 1 (k 1)1k1 k2 k3 ... 1 kk

3.3 Cc nh l ca a thc i xng nhiu bin

nh l 3.3 (nh l tn ti (Theo [2])). Gi s f(x1, x2, ..., xn) l a

thc i xng ca n bin. Khi tn ti a thc (1, 2, ..., n), sao cho

nu vo ch 1, 2, ..., n thay cc biu thc

1 = x1 + x2 + ...+ xn,

2 = x1x2 + ...+ x1xn + x2x3 + ...+ xn1xn,..................

n = x1x2....xn,

th ta nhn c a thc f(x1, x2, ..., xn).

Chng minh. Gi s a.xk11 xk22 ...x

knn l mt n thc ca f(x1, x2, ..., xn).

V f(x1, x2, ..., xn) c tnh i xng, nn cng vi n thc ni trn n

cha qu o ton phn O(xk11 x

k22 ...x

knn ) vi h s a. Nh vy

f(x1, x2, ..., xn) = a.O(xk11 x

k22 ...x

knn ) + f1(x1, x2, ..., xn),

trong f1(x1, x2, ..., xn) l mt a thc i xng c s cc s hng t

hn f(x1, x2, ..., xn). T f1(x1, x2, ..., xn) li c th chn ra mt qu o

b.O(xl11 x

l22 ...x

lnn ) c h s b v v.v... Sau mt s hu hn bc ta c th

phn tch f(x1, x2, ..., xn) thnh tng ca cc qu o ton phn.

f(x1, x2, ..., xn) = a.O(xk11 x

k22 ...x

knn ) + ....+ b.O(x

l11 x

l22 ...x

lnn ).


Theo cng thc (3.2) v (3.4), cc qu o ton phn c biu din qua

cc tng ly tha. Cui cng, trn c s cng thc truy hi (3.5), bng

phng php quy np, cc tng ly tha c biu din qua cc a thc

i xng c s. nh l c chng minh.

Nhn xt. nh l trn cn c th c chng minh bng phng php

s dng khi nim "n thc tri". Tht vy, gi s a.xk11 xk22 ...x

knn l mt

n thc tri nht ca f. Khi k1 k2 ... kn vf1(x1, x2, ..., xn) := f(x1, x2, ..., xn) ak1k21 k2k32 ...knn

l mt a thc i xng c n thc tri thp hn a.xk11 xk22 ...x

knn (nu vo

ch 1, 2, ..., n thay cc biu thc ca chng theo x1, x2, ..., xn. i vi

a thc ny li chn ra n thc tri nht b.xl11 xl22 ...x

lnn v li xt hiu

f2(x1, x2, ..., xn) = f1(x1, x2, ..., xn) bl1l21 l2l32 ...lnn == f(x1, x2, ..., xn) ak1k21 k2k32 ...knn bl1l21 l2l32 ...lnn .

Sau mt s bc hu hn cui cng ta c

0 = f(x1, x2, ..., xn) ak1k21 k2k32 ...knn ... cm1m21 m2m32 ...mnn .T suy ra

f(x1, x2, ..., xn) = ak1k21

k2k32 ...

knn + ...+ c

m1m21

m2m32 ...

mnn .

nh l 3.4 (nh l duy nht (Theo [2])). Nu hai a thc (1, 2, ..., n),

(1, 2, ..., n) sau khi thay

1 = x1 + x2 + ...+ xn, 2 =

1ijnxixj, n = x1x2...xn

cng cho mt a thc i xng f(x1, x2, ..., xn) th

(1, 2, ..., n) (1, 2, ..., n).Chng minh nh l ny hon ton tng t trng hp hai bin.

Biu din mt a thc i xng nhiu bin f(x1, x2, ..., xn) qua a thc

(1, 2, ..., n) ca cc a thc i xng c s theo phng php nh

trong chng minh nh l 3.2 rt phc tp v phi lin tip biu din cc

qu o ca cc n thc nhiu bin qua cc qu o ca cc n thc vi

s bin t hn. V vy khi s bin ln hn 3 tm (1, 2, ..., n) thng

s dng phng php h s bt nh. Xt v d sau y.


V d 3.1 (Theo [5]). Biu din a thc i xng ca cc bin x1, x2, x3, x4sau y theo cc a thc i xng c s 1, 2, 3, 4:

f(x1, x2, x3, x4) = (x1 + x2)(x1 + x3)(x1 + x4)(x2 + x3)(x2 + x4)(x3 + x4).

Li gii. V f c bc 6 nn khi biu din qua cc a thc i xng c s th

ta s c mt tng ca cc n thc dng Ak11 k22

k33

k44 , vi

k1 + 2k2 + 3k3 + 4k4 = 6,

trong k1, k2, k3, k4 l cc s nguyn khng m. Do ta tm f dng:

f(x1, x2, x3, x4) = A161 + A2

412 + A3

313 + A4

21

22+

+A5214 + A6123 + A7

32 + A824 + A9

23, (3.14)

trong

1 = x1 + x2 + x3 + x4, 2 = x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4,

3 = x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4, 4 = x1x2x3x4.

tm cc h s A1, A2, ..., A9 ta cho cc bin x1, x2, x3, x4 nhn cc gi

tr c th no .

Vi x1 = x2 = x3 = 0, x4 = 1, ta c f = 0, 1 = 1, 2 = 3 = 4 = 0 v

t (3.14) suy ra A1 = 0

Vi x1 = x2 = 0, x3 = 1, x4 = 1, ta c f = 0, 1 = 0, 2 = 1,3 = 4 = 0 v t (3.14) suy ra A7 = 0

Vi x1 = x2 = 0; x3 = x4 = 1, ta c f = 0, 1 = 2, 2 = 1, 3 = 0, 4 = 0

v t (3.14) vi ch A1 = A7 = 0 suy ra

4A2 + A4 = 0

Vi x1 = x2 = 0; x3 = 1, x4 = 2, ta c f = 0, 1 = 3, 2 = 2, 3 = 4 = 0

v t (3.14) vi ch A1 = A7 = 0 suy ra

9A2 + 2A4 = 0

Do t hai phng trnh trn suy ra A2 = A4 = 0. Nh vy by gi f c

dng

f(x1, x2, x3, x4) = A3313++A5

214+A6123+A824+A9

23, (3.15)

Vi x1 = 0, x2 = x3 = 1, x4 = 2, ta c f = 4, 1 = 0, 2 = 3,3 = 2, 4 = 0 v t (3.15) suy ra A9 = 1.Vi x1 = 1, x2 = 1, x3 = 1, x4 = 1, ta c f = 0, 1 = 0, 2 = 2,


3 = 0, 4 = 1 v t (3.15) suy ra A8 = 0.

Vi x1 = 0, x2 = 2, x3 = 1, x4 = 2, ta c f = 16, 1 = 3, 2 = 0,3 = 4, 4 = 0 v t (3.15) vi ch A9 = 1 suy ra A3 = 0.Vi x1 = 1, x2 = 1, x3 = x4 = 1, ta c f = 0, 1 = 2, 2 = 0,3 = 2, 4 = 1 v t (3.15) vi ch A3 = 0, A9 = 1 suy ra A5 = 1.Vi x1 = x2 = x3 = 1, x4 = 0, ta c f = 8, 1 = 3, 2 = 3, 3 = 1, 4 = 0

v t (3.15) vi ch A3 = 0, A9 = 1 suy ra A6 = 1.Vy ta c kt qu

f(x1, x2, x3, x4) = 214 + 123 23.

3.4 a thc phn i xng nhiu bin

nh ngha 3.8 (Theo [2]). a thc phn i xng nhiu bin l a thc

thay i du khi thay i v tr ca hai bin bt k.

Chng hn, a thc

(x y)(x z)(x t)(y z)(y t)(z t)l a thc phn i xng theo cc bin x, y, z, t.

a thc

T (x1, x2, ..., xn) =i

f(a, b, c, d) = (b+cad)4(bc)(ad)+(c+abd)4(ca)(bd)++(a+ b c d)4(a b)(c d).

Li gii. Ta thy f l a thc phn i xng. Tht vy, i ch a v b ta

c

f(b, a, c, d) = (a+cbd)4(ac)(bd)+(c+bad)4(cb)(ad)++(b+ a c d)4(b a)(c d) =

= (b+ c a d)4(b c)(a d) (c+ a b d)4(c a)(b d)(a+ b c d)4(a b)(c d) = f(a, b, c, d).

Nh vy khi i ch a v b th f i du. i ch hai bin bt k khc cng

cho kt qu tng t. Vy f l a thc phn i xng theo cc bin a, b, c,

d. Mt khc, f l a thc bc 6, nn theo nh l (3.5) trn, ta c

f(a, b, c, d) = k.T (a, b, c, d) = k.(a b)(a c)(a d)(b c)(b d)(c d),trong k l mt hng s no . xc nh k ta c th cho a, b, c, d

cc gi tr bt k no i mt khc nhau, chng hn a = 0, b = 1, c =

2, d = 3. Khi ta c

f(0, 1, 2, 3) = 192, T (a, b, c, d) = 12

suy ra k=16. Vy ta c kt qu

f(a, b, c, d) = (b+cad)4(bc)(ad)+(c+abd)4(ca)(bd)++(a+ b c d)4(a b)(c d) =

= 16.(a b)(a c)(a d)(b c)(b d)(c d).nh ngha 3.9 (Theo [2]). Bnh phng ca a thc phn i xng n

gin nht T (x1, x2, ..., xn) c gi l bit thc ca cc bin x1, x2, ..., xn:

4(x1, x2, ..., xn) = T 2(x1, x2, ..., xn) =i

Khi s bin ln hn 3, bit thc c biu din qua cc tng ly tha bng

cng thc sau

nh thc sau y c gi l nh thc Vandermonde

Vn =

1 x1 x

21 ... x

n11

1 x2 x22 ... x

n12

... ... ... ... ...1 xn x

2n ... x

n1n

Khi ta c

4(x1, x2, ..., xn) =

ijn(xi xj)2 = V 2n .

Cng thc trn cn c th c vit nh sau

(x1, x2, ..., xn) =

1 1 1 ... 1x1 x2 x3 ... xn... ... ... ... ...xn1

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