chuyên đề toan thpt

8/14/2019 chuyn toan thpt

1/24

TRNG I HC S PHM H NIKHOA TON TIN

*********

Chuyn

QUI NP TON HC

Gio vin hng dn: ng nh HanhSinh vin thc hin: Nguyn Ngc Th

Lp: HK53Ton

H NI,THNG 11-2006


2/24

Chuyn : Qui np ton hc

I DUNG CHNH

1. Phng php gii2. Cc dng ton in hnh3. V d minh ho4. Li gii chi tit 5. Ch 6. Bnh lun phn tch7. Bi tp

Nguyn Ngc Th - Lp HK53 Ton. 2


3/24


Li m u

Trong khun kh gii hn ca mt chuyn nhm bin son chng ti xinkhng a ra cc khi nim nh ngha,mnh , nh l v cc tnh cht ctrong SGH ph thng m ch a ra cc dng ton km theo phng php gii ,tip l cc v d minh ha cng li gii chi tit. Kt thc v d l nhng ch cn thit nhm tng cht lng s phm cho chuyn . Sau mi dng ton chnti c a ra mt lot cc bi tp ngh cc bn tham kho v th sc.

Khi no cn dng n kin thc no chng ti s vn trnh by li trc kh s dng trong bi gii ca mnh. Mc d tham kho mt lng rt ln cc tliu hin nay cng vi s n lc ca bn thn nhng do trnh hiu bit c hnn chc chn khng trnh khi thiu st rt mong c s gp ca thy gio ng nh Hanh v tp th lp K 53H. Xin chn thnh cm n.

Mt ln na nhm bin son chng ti xin by t lng cm n ti thy gio ng nh Hanh c v, ng vin, gi , trong qu trnh chng ti thc hin

chuyn ny, chuyn sm c hon thnh. Xin chn thnh cm n Thy.Chng ti cng xin chn thnh cm n bn Phm Tr My cung cp cho

chng ti nhiu ti liu hay v qu trong qu trnh thc hin chuyn . Th gp ca cc bn xin gi v a ch email : [email protected]

Ngi thc hin

Nguyn Ngc Th.



4/24


QUI NP

Phng php qui np thc s c hiu lc vi lp cc bi ton chng minh mmnh ph thuc vo s t nhin n N.

Phng php gii

chng minh mt mnh Q(n) ng vi min p , ta thc hin 2 bc theoth t: Bc 1: Kim tra mnh l ng vin p

Bc 2: Gi s mnh ng vin k p , ta phi chng minh rng mnh ng vi 1n k .

Cc dng ton minh ho.

Dng 1 : Dng phng php qui np chng minh mt ng thc . VD1: Chng minh rng : vi mi s t nhin n 2 ,ta c :

an bn = (a b)(an 1 + a n 2.b + +a.bn -2 +b n 1) (1) Ta chng minh ng thc (1) bng phng php qui np.

Gii

Khi n=2 th VT(1) = a2 b 2 , VP(1) = (a b)(a+ b)= a2 b2 .

Vy ng thc (1) ng vi n=2.

Gi s (1) ng vi mi n = k 2 , tc l :

a k b k = (a b )(ak-1 + a k-2.b + + a.bk-2 + b k-1 )Ta CM (1) cng ng vi n=k + 1 , tc l :



5/24


a k+1 b k+1 = (a-b)(ak + a k-1.b ++ a.bk-1 + bk )

Tht vy : p dng gi thit qui np , ta c : a k+1- b k+1 = a k+1 ak .b+ak .b bk+1

= ak (a-b) + b(ak -bk )

= ak (a-b) +b(a-b)(ak-1 + a k-2.b + + a.bk-2 + b k-1 )

= (a-b) [ak + b(ak-1 +a k-2.b ++a.bk-2 +b k-1) ]

= (a-b)(ak +a k-1.b ++a.bk-1 +bk )

Vy (1) ng vi mi s t nhin n 2.

Bnh lun: Trong li gii trn ta dng k thut thm bt s hng bc chng minh (1)ng vi n = k+1 ,lm nh vy ta s dng c gi thit qui np ca bi ton. y l mt k thut hay c hiu lc mnh m trong vic n gin ho li gii, c dng rng ri trong qu trnh gii nhiu dng ton khc nhau ng vi nhiu chuyn khc nhau ca ton ph thng . V d sau cho thy r iu ny.

(TTS_khi A2002cu1 )

Cho phng trnh : 0121loglog 2323 =++ m x x (2) ( m l tham s )

1. Gii phng trnh (2) khi m = 22. Tm m phng trnh (2) c t nht mt nghim thuc on33;1 . Bnh thng nu khng dng k thut thm bt th nhiu hc sinh s lm nh sau :

iu kin 0> x , t 23log 0t x , khi pt (2) vn l dng v t ,tt nhin vic gii(2) khng c g kh khn sau mt hi lus cho ta p n . Tuy nhin nu ta thm 2ng



6/24

Chuyn : Qui np ton hcthi bt i 2 vo v tri ca phng trnh (2) th li mt ng cp khc . Khi phntrnh (2) tr thnh :

0221log1log 2323

=+++ m x x

iu kin 0> x . t 11log23 += xt ta c : 022

2=+ mt t (3) . R rng (3) l

phng trinh bc 2 i vi bin t, vic gii (3) n gin v nhanh hn nhiu so vi gi phng trnh m cch t u tin mang li . Cng phi ni thm rng vn c hc sinhmay mn thy trong phng trnh c s gp mt ca cn thc lp tc t t bng cnthc v dn ti pt(3) nh trn. Nhng ch l may mn ngoi l m mt s t bi tonmang li trong phi k n bi ton trn.

Qua phn tch v d trn ta thy li ch v s hiu qu m k thut thm bt em licho chng ta trong vic gii ton ph thng l rt ln.

Ta s gp li k tht ny trong li gii v d (5) ngay sau y v mt s v d khc na cmt trong chuyn ny .Xin mi cc bn cng theo di. VD2: CMR: Mi s t nhin n 1 , ta c :

( ) ( )( )6

1211...321 2

2222 ++=+++++nnn

nn

(2)

Gii

Khi n = 1 VT(2) = VP(2) nn (2) ng.

Gi s (2) ng vi n = k 1 , tc l :

( )6

)12)(1(1...321. 22222

++=+++++ k k k k k

Ta phi chng minh (2) cng ng vi n = k +1 , tc l :

( )[ ] ( )6

)3)(2)(1(111...321 22222 +++=+++++++ k k k k k

Tht vy : 12+22+32++(k-1)2+ k 2 +(k+1)2

= 2 22 2 2 21 2 3 ... 1 1k k k



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6)12)(1( ++= k k k

+ (k+1)2

= 22 7 61

6k k

k

6)32)(2)(1( +++= k k k

.Vy (1) ng vi mi s t nhin n 1.

Ch : li gii trn khng c g c bit ngoi k nng nhm s hng tinh t thnh lp s xut hin ca gi thit qui np bc n = k+1 dn n gii quy

bi ton.

VD3 Tm s hng tng qut ca dy s sau :

nn uuu 2,3 11 == + , ( )1n

GiiTa c :

011

2 12

3 2

1

3 3.22. 2.3 6 3.22. 2.6 12 3.2

.........................................3.2nn

uu uu u

u

Ta s chng minh 13.2nnu (3) bng qui np .

Khi n = 1 ta c 1 3u ( ) dogt (3) ng .

Gi s (3) ng vi n = k,( )1k

tc l :1

3.2k

k uTa phi chng minh (3) ng vi n = k+1 , tc l :1 3.2k k u

Tht vy : 11 2. 3.2.2 3.2k k k k u u Vy (3) ng vi n = k+1 nn cng ng vi mi1n .

Ch : Sau v d ba ta rt ra phng php gii chung cho dng ton tm s hng tng qut ca mt dy s gm hai bc :



8/24

Chuyn : Qui np ton hc Bc 1:Tm vi s hng u ca dy Bc 2 : D on s hng tng qut, ri chng minh bng qui np.

VD4: Tnh o hm cp n ca hm s sau : x y += 11

Gii

Ta c :2

,

)1(1 x

y+= ,

3,,

)1(2.1 x

y+= ,

4,,,

)1(3.2.1

x y

+=

,,)(n y

By gi ta tm)(n y bng quy np nh sau :

Gi s( ) ( )

( ) 11!1++

= k k k

xk y

Ta c :( )[ ] 1)1(

1

)1(2

,)()1(

)1()!1()1(

)1()1)(1)(1(

.!1 +++

++

++=

+++== k

k

k

k k k k

xk

x xk

k y y

Vy1)1(

!).1(++

= nn xn

y

Ch : Phng php gii chung cho dng ton ny c th phn lm hai bc nh sau :

Bc 1: Tnh o hm cp mt , hai,ba,,cho ti khi d on c ohm cp n. Bc 2: Chng minh o hm cp n ng bngqui np ton hc .

VD5 : ( thi hc k 1, i s tuyn tnh - lp K53GH_2003)CMR : Nu s phc z tha mn :

cos21

cos21 =+=+ nn z z z z (5)

Gii

Vi n=1, z

z VT += 1)5(

, VP(5)= cos2 theo gi thit (5) ng .



9/24

Chuyn : Qui np ton hcGi s (5) ng vi n=k , tc l :1 2cosk k z k z

Ta phi chng minh (5) cng ng vi n=k+1, tc l : 111 2cos 1k k z k z

Tht vy : 1 11 11 1 1 1k k k k k k z z z z z z z z

( ) .1cos2cos2.cos2 = k k

1. 4 cos( 1) cos( 1) 2cos 12

k k k

=2cos(k+1)

Vy (5) ng vi n = k +1,nn (5) ng vi1 n .

Ch : khng bnh lun thm v li gii trn . Tht bt ng khi y li l thihc k cp i hc . iu ny chng t qui np khng phi mt vn ngulnh trong cc k thi.Do vic nm vng phng php gii l iu tht cn thivi mi ngi hc v lm ton.

Bnh lun chung cho dng mt : Qua nm v d trn ta thy bi ton chng minh ng thc bng cch dng phng php qui np ton hc ch kh khn v phc tp phn cu

bc 2 , tc l chng minh ng thc ng vi n=k+1.Khi t ng thc cn chngminh ng vi n=k+1,ta bin i kho lo,(dng k thut thm bt ,hoc tch s hng ) s dng c gi thit ng thc ng vi n=k,tip tc thc hin tnh ton mt s bna ta s c pcm.

Cn nhm mnh rng vi dng ton mt ta thng bin i theo con ng ny ! Tuy nhin y khng phi l cch bin i duy nht,ta c th bin i trc tip t gi thit ng thc ng vi n = k (gi thit qui np ca bi ton) , suy ng thc ng vi n =k+1. minh ho cho cch lm ny ta cng nhau i xt v d sau y :

CMR mi n thuc N* ta c : nnnn

3.432

43

3...32

31

2

+=+++ (BL)

Gii

Vi n = 1 , th (BL) :1 3 53 4 12

ng.

Gi s (BL) ng vi n = k, tc l : 21 2 3 2 3...3 3 3 4 4.3k k

k k (BL.1)



10/24


Ta phi chng minh (BL) ng vi n = k+1, tc l:

( ) ( )112 3.4

312

4

3

3

1

3...3

2

3

1++

++

=

+

++++ k k k k k k

(BL7.2)

Tht vy : Cng vo hai v ca (BL7.1) mt lng l : 11

3k k

, ta s c (BL7.2)

Vy (BL) ng vi n = k+1, nn cng ng vi mi n thuc N* .K thut bin i ny s mt ln na c th hin v d (8) trong dng hai qui npton hc. Xin mi cc bn cng theo di.

Bi tp ngh.

Bi 1: CMR : Mi n* N , ta c : 1+3+5++(2n-1) = n2

Bi 2 : CMR: * N n , ta c : 11 2 3 ...2

n nn

Bi 3: CMR : M i *n N ,ta c : ( )4

1...21

2333 +=+++ nnn

Bi 4: CMR : Mi a >0, a 1, 1 2, ,..., 0n x x x ,ta c h thc sau:

1 2 1 2... ...log log log log n n x x x x x xa a a a Bi 5: CMR: Mi s t nhin n 1, vi mi cp s (a,b),ta c cng thc sau y, gi lcng thc khai trin nh thcniutn. n 0 n 1 n-1 1 2 n-2 2 k n-k k n n= + + +...+ +...+(a+b) C a C a b C a b C a b C bn n n n n

Bi 6: CMR : 213 3 3 31 2 3 ...

2

n nn sn

Bi 7: CMR: Vi mi s t nhin n 1,ta c ng thc :

2)1(

...321+=++++ nnn

Bi 8: CMR : mi n thuc N ta c :( ) ( ) nn

n 2121

2

11...

254

194

114

1 2 +=

Bi 9: Tnh o hm cp n ca cc hm s sau :

a) )1ln( x y += b) 1

1 y

x x



11/24

Chuyn : Qui np ton hcc) ax y sin= a const d) 2sin y x

Bi 10: Tm tng s )1(1

...3.2

12.1

1nn sn

Bi 11:Tm s hng tng qut ca cc dy s sau :a) 1 1

13, 2 .2n n

u u u

b) 1 1, .n nu a u a bu Cc bi tp ngh chng ti a ra c la chn cn thn, k lng, phnno c tnh cht nh hng phn loi theo cc loi ton cha trong dng mt .

Dng 2: Dng phng php qui np chng minh mt bt ng thc.VD1: Chng minh bt ng thc Bec-nu-li(Bernoulli). Nu h >0 , vi mi s t

nhin n 2 nhh n +>+ 1)1( (1) ,

Gii Nu n =2, ta c : (1+h)2 = 1+2h+h2 > 1+2h (v h2> 0) .Vy (1) ng .Gi s (1) ng n n = k , tc l :( 1+h)k > 1+kh (2).Ta phi chng minh (1) cng ng n n =k+1 ,tc l : (1+h)k+1 > 1+(k+1)h.

Tht vy : (1+h)k+1 =(1+h)(1+h)k ( 2)do

(1+h)(1+kh) =1+h+kh+kh2

= 1+h(1+k)+kh2 > 1+h(1+k).(v kh2 >0)Vy (1) ng vi mi s t nhin n 2.

Ch : Php chng minh trn gi thit h khng ph thuc n . Trong trng hp h ph thuc n , ngi ta chng minh rng bt ng thc bec_nu_li vn ng (dngcng thc nh thc niutn ) .

VD2 : ( 101 cu 4a_BTTS)Chng minh rng nu x >0 th vi mi s t nhin n ta u c :

!...

2!x

x12

n x

en

x ++++>(2)

Gii



12/24


Xt hm s . 2

1 ...2! !

n x

n

x x x e x

n f

Ta

phi chng minh : 0, : 0n x n N x f (2.1)Tht vy , ta c : , 0 0nn f Xt 1 1

x x e x f Ta c

,

11 0, 0 , x x e x f ( ) x f 1 tng vi mi x >0 1 1 0 x f f

Vy cng thc (2.1) ng vi n=1.

Gi s bt ng thc ng vi n=k.

Ta c: 2

0, 1 ... 02! !

k x

k

x x x x e xk

f

(2.2)

Ta phi chng minh :

2 1

10, 1 ... 0

2! ! 1 !

k k x

k

x x x x x e x

k k f

Tht vy , ta c :

1,

1

12 .1 ...2! ! 1 !

k k x

k

k x x k x x e

k k f

1,

11 ...

1 !

k k xk k

x x x e x xk k f f

Theo (2.2) c( ) ( ) ( ) x x x f f f

k k k 1

,

100

++>>

tng vi

( ) ( ) 00011

=>>++ f f k k x x

Vy bt ng thc ng vi n=k+1 nn cng ng vi mi s t nhin n .

Ch : Nhn vo bt (2) ta thy c hai v u l cc hm s ca bin x . Nu ta

chuyn ton b v phi ca bt (2) sang v tri v t bng ( ) x f n bi ton tr

thnh Cmr : ( ) 0,,0 >> x N n x f n . Khi dng qui np x l bi ton kt

hp vi ng dng ca o hm v tnh n iu ca hm s l v cng hp l.Rrng im mu cht,bc t ph a n hng gii p cho bi ton l thao tcchuyn v .



13/24


VD3 (131CU4a_BTTS):Cho hm s f xc nh vi mi x v tho mn iu kin :

f(x+y) f(x).f(y) vi mi x,y (3)CMR : Vi mi s thc x v mi s t nhin n ta c :

( )

n

n

x f x f

2

2

(3.1)

Gii

Trong BT f(x+y) f(x).f(y) thay x v y bng2 x

, ta c:

( ) ( )

2

22.

222 +

x f x f x f x f x x f

Vy bt ng thc( )

n

n

x f x f

2

2

ng vi n=1

Gi s bt ng thc ng vi n =k , 1k

. Ta c( )

k

k

x f x f

2

2

Ta chng minh bt ng thc ng vi n = k+1, tc l :( )

12

12

+

+

k

k

x

f x f

Tht vy ta c :2

1 1 12 2 2 2

22 2

12 2

12 2

12 2

x x x x f f f k k k k

k k x x f f k k

k k x x f f k k

Do tnh cht bc cu ta c c :( )

12

12

+

+

k

k

x f x f

Bt ng thc ng vi n = k+1 nn cng ng vi mi s t nhin n.



14/24


15/24

Chuyn : Qui np ton hcTa phi chng minh : 2 1k k u u

Ta c :( 5.1)

12 1

1 12 2

dok k

k k u uu u

Vy (5) ng vi n = k+1 nn cng ng vi mi n thuc N* .Chng minh dy cho l b chn di. Ta dng qui np chng minh :

*1,u n N n (6)Khi n=1 , 2 11u nn (6) ng.

Gi s (6) ng vi n = k , 1k ngha l 1uk (6.1)

Ta phi chng minh : 1 1k u

Ta c : 1211

21

1 =+

>+

=+k

k

uu . Vy 1 1k u .Dy s cho b chn di bi 1.

Ch : Khi gp dng ton chng minh dy s n iu v b chn ta thc hinnh sau :

bc 1 : Dng qui np chng minh dy s l n iubc 2 : D on s M trong trng hp dy b chn trn bi M v S mtrong trng hp ngc li.Sau dng qui np chng minh dy b chnbi trn bi M hoc b chn di bi m trong trng hp ngc li .

VD 6:

Chng minh rng : 2,,11 ><

+ n N nn

n

n

(6)

Gii

Khi n =3 bt (6) tr thnh 32764

31

13


16/24

Chuyn : Qui np ton hc Ch : li gii trn ta dng phng php lm tri nh gi ca bt bc n

=k +1,ti v tr du bt (2).C th ni y l phng php ch cng, mang tnhc th trong chng minh bt .Hc sinh cn nm vng v lm tt phng php nyv s hiu qu m n mang li, cng lu rng khng nn nh gi bt qu lng ,hoc qu cht . Sau y l mt v d minh ho na ginh cho phng php nh gi lm tri.

VD 7: Cho x1,x2,,xn l cc s dng. Chng minh rng :

4,2...3112

1

2413

2

2

1 ++++++++++ n

x x x

x x x

x x x

x x x

x x x

n

n

nn

n

n(7)

GiiVi n = 4 , bt c dng :

2231

42

42

31

31

4

24

3

13

2

42

1 ++++

++++++++ x x

x x x x x x

x x x

x x x

x x x

x x x ng.

Gi s bt(7) ng vi n = k . Tc l :

( )4,2...112

1

13

2

2

1 ++++++++ k

x x x

x x x

x x x

x x x

k

k

k k

k

k (7.2)

Ta chng minh bt(7) ng vi n = k+1.

Do vai tr bnh ng gi cc xi ( i = 1,2,,k+1), nn khng gim tnh tng qut ca biton ta c th gi s xk+1= min{ x1,x2,,xn } , tc l : 0, ,1 1 1 1 x x x x xk k k k Do vy ta c :

1113

2

2

1

1

1

1113

2

12

11

......

+

+++ ++++++>++++++++= k

k

k k

k

k k

k

k k x x

x x x

x x x

x x x

x x x

x x x

x x x

x s

(7.1)

Do: 0;;1

1

11112

1

12

1 >++

++

+

+

++ k

k

k

k

k k

k

k k x x

x

x x

x

x x

x

x x

x

x x

x (7.3)

T (7.1),(7.2),(7.3) suy ra 21 >+ sk . Vy bt ng vi 1n k nn cng ng vi mi n . l pcm.

Ch : Th d trn cng cho thy r nt sc mnh ca phng php nh gi lmtri trong chng minh bt .Bc ngot a n hng gii quyt cho li gii bton l thao tc nh gi , c lng , gi tr ca xk+1 = minxi ,{ i= 1,2,n} bc n = k+1 .



17/24

Chuyn : Qui np ton hcVD 8 : Chng minh rng : 1n , ta c ( )

12

12.6.4.2

12...5.3.1

+


18/24

Chuyn : Qui np ton hcChng minh rng :tgn ntg

Bi 2Chng minh rng : vi a >0 th 1 4 1...2aa a a

Bi 3 Chng minh rng : 1 1 , 3nn n nn Bi 4 Chng minh rng vi mi s t nhin n ta c :

121

...3

1

2

11)

1....321)

++++

+++++

nn

b

nn

nna

c)212

1...

31

21

1 2n>

++++

Bi 5 Chng minh bt ng thc :

( )( )( ) ( ) 12232222 2.3

121...212121

+


19/24

Chuyn : Qui np ton hcBi 13 Cho n s dng nghim ng iu kin .1.... 21 =naaa CMR :

(*)...21 naaa n +++

Du =xy ra khi no ?

Bi 14 Chng minh mi s t nhin n >1, ta c : 1 cos cos 11

n nn n

Bi 15 Cho n l s t nhin v 0 1 2n

CMR : ( )( ) tg tgnnn .cos1cos1 ++

++>

nnnCMRn N n

Qua hai dng u ca qui np ton hc ta c cm gic mc hay va kh ca biton tng dn.Do c th ca n ,hai dng ny c hc tng i su ph thng.Dng ba ca bi ton ,cng l dng cui cng chng ti s trnh by trong chuyn ny c hc s qua bc ph thng v hc cao hn nm th haica trng hsp... Cng v l do m dng ba c chng ti a vo sau cng . Xin mi cc bn chuyn sang dng ba ca qui np ton hc.

Dng 3: Dng qui np ton hc chng minh mt biu thc dngUn chia ht cho mt s t nhin .

VD1: Chng minh rng nnna N n n 53, 23* ++= chia ht cho 3 . (1)

GiiVi n = 1 ta c : 391.51.313 21 =++=a ng .

Gi s (1) ng vi n = k ,( )1k , tc l : 353 23 k k k a k ++=

Ta phi chng minh (1) ng vi n = k+1, ngha l : ( ) ( ) ( )315131 231 +++++=+ k k k a k

Tht vy : 55363133 2231 ++++++++=+ k k k k k k a k

3993533

2

3

23 +++++= k k k k k

Vy (1) ng vi n = k+1, nn cng ng vi mi* N n

Ch : Ta bit rng mt tng chia ht cho mt s khi tng s hng ca tng chia

ht cho s . Nhn thy1+k a l mt tng cc a thc ca k , Vy chng minhak+1 chia ht cho 3 ta phi thc trin ak+1 , sau tin hnh thc hin sp xp li



20/24

Chuyn : Qui np ton hccc s hng , kt hp vi gi thit qui np , vit li ak+1di dng tng cc s hng chia ht cho 3.

VD2:Chng minh rng 2 n , ta c : an = ( )( ) ( ) nnnnn 2...21 +++ (2)

GiiKhi n = 2 , ta c : a2 = ( )( ) 222212 ++ ng

Gi s (2) ng vi n =k , 2k , tc l : ak = ( )( ) ( ) k k k k k 2...21 +++

Ta phi chng minh (2) ng vi n = k+1, ngha l :ak+1 = ( )( ) ( ) 1211...2111 ++++++++ k k k k k

( )( ) ( ) ( )( ) ( )( )( )21...322...32 +++++++=++++= k k k k k k k k k k k k

11 2 3 ... .2. 1 222

k k k k k k k k

k 1 4 2 4 31 4 4 4 4 4 2 4 4 4 4 4 3

Vy (2) ng vi n = k+1 ,nn (2) ng vi 2 n . Ch : Li gii v d hai khng c g mi l , ta thc hin k thut vit li ak+1 ,

thnh lp s xut hin gi thit qui np , d dng suy ra pcm.

VD3: Chng minh rng : an = 1,67627263 33 + nnn (3)

GiiVi n = 1 , ta c :a1 = 676676271.263 31.3 =+ nn (3) ng.Gi s (3) ng vi n = k , 1k tc l : ak = 67627263 33 + k k (3.1)Ta phi chng minh (3) ng vi n = k+1, tc l : ak+1= ( ) ( ) 676271.263 313 +++ k k

Tht vy :ak+1 = ( ) 676)1.3(676333)1(3 67667627263.2727)1(263 ++=+ +++ k k k

do

k k

Vy (3) ng

vi n = k+1 , nn (3) ng vi mi .1n

Ch : v d ny vic vit li ak+1 khng ch n thun l s thc trin sp xpli cc s hng , r rng y k thut thm bt li pht huy tc dng, vic a 27ra ngoi lm tha s chung, vi mc ch thnh lp c gtqn lm d ra mt lng so vi lng ban u , cn bng bi ton ta thm vo mt lng 676k+676 . Lm nh vy ta s dng c gtqn tin n kt thc li gii.

V d bn di y l mt minh hc na cho li gii loi bi tp ny.



21/24


VD4: Chng minh rng: 382.32.5:1 121112 ++ + nnnnn (4)Vi n = 1 , ta c : 38382.94.52.32.5 11.2111111.2 =+=+ ++ nn (4) ngGi s (4) ng vi n = k,( )1k tc l : 382.32.5 121112 ++ + k k k k (4.1)Ta phi chng minh (4) ng vi n = k+1, tc l :

382.32.5 1221111122 ++++++ + k k k k

Tht vy :

( ) 382.3.382.32.5.50

23.122.5.502.4.3.32.2.5.252.32.5

38

121

)1.4(38

121112

1211121211121221111122

+++

++++++++++

+=

+=+=+

k k

do

k k k k

k k k k k k k k k k k k

Vy (4) ng vi n = k+1 , nn cng ng vi * N n .

VD5: Chng minh rng : 1 n , ta c : 241021143 234 nnnn + (5)Gii

Vi n = 1 ta c : 2401021143 =+ , nn (5) ng .Gi s (5) ng vi n = k, 1k , ngha l : 241021143 234 k k k k + (5.1)

Ta phi chng minh (5) ng vi n = k+1 , ngha l :( ) ( ) ( ) ( ) 2411012111413 234 +++++ k k k k

Tht vy :

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )2411.1210 _ 211431101221

133141464311012111413

24)1.5(24

2342

2234234

++=+++++++++++=+++++

k k k k k k k k k k

k k k k k k k k k k k

do

Vy (5) ng vi n = k+1 , nn cng ng vi1 n .

Ch : V d 5 v v d 1 thuc cng mt dng .Do cch gii ginh cho v d 5 xem ch v d 1.

Bnh lun chung cho dng 3:Qua nm v d ginh cho dng ba ta thy mucht gii tt cc bi tp ca dng ba l k nng vit li an ng vi n =k+1,thnh tng cc s hng hoc tch ca cc tha s chia ht cho s t nhin cn chng minh . Tt nhin trong qu trnh vit li nh vy, ta vn lu



22/24

Chuyn : Qui np ton hc ti vic s dng gi thit qui np ca bi ton.C th ni k thut vit l ca mt bi ton ni ring v vit li mt biu thc ton hc ni chung dng c gi thit ca bi ton , c bit c hiu qu, trong gii ton ph thng. Xin a ra mt s v d in hnh cho k thut ny.

V d 1 (TTS_khiA2003cu1 )

Gii h phng trnh3

1 1

2 1

x y x y

y x

iu kin 0 xy . H cho c vit li d dng :

32 1

x y x y

xy

y x

Nh k thut vit li , ta xc nh c hng gii cho h trn l xu pht t phng trnh th nht ca h .V d 2(HCSNN_khi A2000)

Cho h phng trnh : 2

1

x xy y m

xy x y m

1. Gii h cho khi m=-3.2. Xc nh m h c nghim duy nht.

H cho c vit li di dng : 2

1

x y xy m

xy x y m

Nh vit li h nh vy m ta c th t x+y= S, xy = P , iu kin S 2-4P 0

, khi vic gii h pt trn khng c g c khn. Ni chung y l k thut c bn trong gii ton , hc sinh nn rn luyn kthut ny c th p dng trong qu trnh gii tt c cc dng ca ton h s cp.

Bi tp ngh.

Bi 1: CMR 22511516:* n N n n

Bi 2: CMR , 13 1 6nnn N u Bi 3: CMR 2 1 2,12 11 133n nn N

Bi 4: CMR 6436323.4: 22 + + n N n n



23/24

Chuyn : Qui np ton hcBi 5: CMR nn N n 2: 3 + chia ht cho 3

Bi 6: CMR 1 2 ... 2n n n chia ht cho 1.3.5... 2 1 ,n n N .

Ti liu tham kho

1. Don Minh Cng : Gii thiu thi tuyn sinh i hc 2000-2001,NXBGD.

2. H Vn Chng : Tuyn tp 700 bi ton bt ng thc,NXB.Tr1998

3. Phan c Chnh V Dng Thy _o Tam L Thng Nht : Cc b ging luyn thi mn ton ,tp hai NXBGD

4. ng nh Hanh :Tp bi tp quan h chia ht ginh cho K53GH.

5. Trn Vn K : 460 bi ton bt ng thc,NXB.Tr TPHCM.

6. Ng Thc Lanh- V Tun- Ng Xun Sn : i s v gii tch 11, Nh xut bn gio dc1998

7. V i Mau :Phng php gii ton bt ng thc,NXB.Tr2000

8. Nguyn Vn Mu :Mt s bi ton chn lc v dy s ,NXNGD

9. Trn Phng : Phng php mi gii thi tuyn sinh mn ton, NXBGD.

10.Nguyn Tin Quang :Bi tp s hc, NXBGD.

11. B thi tuyn sinh i hc, Nh xut bn gio dc 2001..

12.Tuyn tp 30 nm tp ch ton hc v tui tr, Nh xut bn gio dc.



24/24

Chuyn : Qui np ton hc13.Hai cun sch gio khoa chnh l hp nht nm 2000 l i s v gii tch

11, gii tch 12 ca Nh xut bn gio dc .


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