tai lieu bai tap toan cao cap a2 c2 docx

B CNG THNGTRNG I HC CNG NGHIP THC PHM

TP. HCM

L HU K SN

Bi tpTon cao cp A2 - C2

MSSV: . . . . . . . . . . . . . . . . . . . . . . . . . .H tn: . . . . . . . . . . . . . . . . . . . . . . . . . .

TP. HCM Ngy 15 thng 2 nm 2012

Mc lc

1 MA TRN V NH THC 31.1 Ma trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 nh thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Ma trn nghch o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Hng ca ma trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 H PHNG TRNH TUYN TNH 8

3 KHNG GIAN VECTOR 93.1 Khng gian vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Khng gian Euclide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 NH X TUYN TNH 11

4.1 nh x tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Gi tr ring - vector ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 DNG TON PHNG 14

Ti liu tham kho 15

2

Chng 1

MA TRN V NH THC

1.1 Ma trn

1. Cho A =

1 21 33 4

;B = 0 13 22 3

;C =2 31 2

4 1

.Tnh (A+B) + C;A+ (B + C); 3A 2B; (3A)t; (3A 2B)t.

2. Cho ma trn A =

1 2 10 1 23 1 1

;B = 2 3 11 1 0

1 2 1

;C =2 3 01 2 4

4 1 0

.Tnh A.B.C v A.C +B.C.

3. Tnh A =

a b cc b a1 1 1

1 a c1 b b1 c a

.4. Cho ma trn

(1 02 1

), hy tm ma trn A2012.

5. Cho ma trn

(1 05 1


6. Cho ma trn A =

(cos sinsin cos


7. Cho ma trn A =

(0 11 0


8. Cho ma trn A =

(0 01 0

). Tnh ma trn (I A)2012.

9. Cho ma trn J =

0 0 11 0 00 1 0

. Tnh ma trn J201210. Cho ma trn A =

(0 01 0

). Hy tnh tng sau

2012n=0

2nAn = In + 2A+ 4A2 + 8A3 + 16A4 + + 22011A2011 + 22012A2012

3

11. Cho ma trn A =

(0 01 0

). Hy tnh tng sau

2012n=0

An = In + A+ A2 + A3 + A4 + + A2011 + A2012

12. Cho ma trn A =

(0 10 0

). Hy tnh tng sau

2012n=0

2nAn = In + 2A+ 4A2 + 8A3 + 16A4 + + 22011A2011 + 22012A2012

13. Cho ma trn A =

(0 10 0

). Hy tnh tng sau

2012n=0

An = In + A+ A2 + A3 + A4 + + A2011 + A2012

14. Cho ma trn A =

0 1 10 0 10 0 0

. Hy tnh tng sau2012n=0

(2)nAn = In 2A+ 4A2 8A3 + 16A4 + + (2)2011A2011 + (2)2012A2012

15. Cho ma trn A =

(a bc d

), hy tnh A2 (a+ d)A+ (ad bc)I2.

16. Tm f(A) nu

a. f(x) = x2 5x+ 3 vi A =(

2 13 3

);

b. f(x) = x2 x 1 vi A =2 1 13 1 2

1 1 0.

.17. Cho A l ma trn vung cp 1000 m phn t dng i l i. Tm phn t dng 1 ct

3 ca ma trn A2.

18. Cho A l ma trn vung cp 1000 m phn t dng i l (1)ii. Tm phn t dng2 ct 3 ca ma trn A2.

19. Cho A l ma trn vung cp 1000 m phn t dng i ct j l (1)i+j. Tm phn t dng 1 ct 2 ca ma trn A2.

20. Cho A l ma trn vung cp 1000 m phn t dng i l 2i1. Tm phn t dng 2ct 4 ca ma trn A2.

4

21. Hy tm s n nguyn dng nh nht ma trn An = 0 (ma trn-khng), vi

a. A =

0 1 00 0 10 0 0

b. A =0 1 10 0 1

0 0 0

c. A =0 0 10 0 0

0 0 0

d. A =

0 0 1 10 0 0 10 0 0 00 0 0 0

e. A = 0 0 01 0 01 1 0

1.2 nh thc

1. Bit cc s 204, 527, 255 chia ht cho 17. Khng tnh nh thc, chng minh rng:2 0 45 2 72 5 5

chia ht cho 17.2. Tnh cc nh thc sau

1 =

5 3 21 2 47 3 6

; 2 =

1 1 11 0 11 1 0

; 3 =a a aa a xa a x

; 4 =1 1 11 2 31 3 6

5 =

0 1 11 0 11 1 0

; 6 =a b cb c ac a b

; 7 =0 a 0b c d0 c 0

; 8 =a x xx b xx x c

;9 =

a+ x x xx b+ x xx x c+ x

; 10 =sin a cos a 1sin b cos b 1sin c cos c 1

; 11 =

1 1 1x y zx2 y2 z2

; 12 =x y x+ yx x+ y x

x+ y y y

3. Gii cc phng trnh v bt phng trnh

a.

x x+ 1 x+ 2

x+ 3 x+ 4 x+ 5x+ 6 x+ 7 x+ 8

= 0;b.

2 x+ 2 11 1 25 3 x

0;4. Chng minh rng

a.

a1 + b1x a1x+ b1 c1a2 + b2x a2x+ b2 c2a3 + b3x a3x+ b3 c3

= (1 x2)a1 +b1 c1a2 b2 c2a3 b3 c3

;b.

1 a a3

1 b b3

1 c c3

= (a+ b+ c)1 a a2

1 b b2

1 c c2

;5. Hy tnh cc nh thc sau

1 =

4 5 2 62 2 1 36 3 3 94 1 5 6

; 2 =3 9 4 21 2 0 32 3 0 12 1 2 1

; 3 =1 1 1 11 1 2 21 1 1 31 1 1 1

5

6. Hy tnh nh thc ca ma trn

2 b 2 2 bb 2 b2 + 4 4b2 b 4b b2 + 4

p s : nh thc ma trn bng 0.

7. Tnh nh thc cp n: Dn =

5 3 0 0 0 02 5 3 0 0 00 2 5 3 0 0 0 0 0 0 2 5

8. Tnh nh thc Vandermond: Dn =

1 x1 x

21 xn11

1 x2 x22 xn12

1 xn x

2n xn1n

1.3 Ma trn nghch o

1. Tm s thc m ma trn A =

(m 10 m 1

)(m 1 0

1 m 1)(

m 1 01 m 2

)kh

nghch.

2. Cho ma trn A =

0 1 0 00 m 1 00 m m2 14 0 0 0

. Hy tm phn t dng 1 ct 4 ca A1.p s :

14.

3. Cho ma trn A =

(1 23 4

)v B =

(1 2 33 2 1

). Tm ma trn X tha AX = B.

4. Cho ma trn A =

(1 23 4

)v B =

(7 7 11 7 7

). Tm ma trn X tha AX = B.

5. Tm ma trn X tha mn phng trnh

(2 13 2

)X

(3 25 3

)=

(2 43 1

)


1 2 32 6 51 3 2

X =1 02 1

0 1

7. Tm ma trn X tha mn phng trnh X

1 2 02 2 31 1 1

= (2 1 00 1 1

)

8. Tm ma trnX tha mn phng trnh

3 0 18 1 15 3 2

1 1 11 0 11 1 2

X =3 0 18 1 1

5 3 2

9. Tm ma trnX tha mn phng trnhX

1 2 13 2 02 3 1

2 3 50 1 62 0 6

=2 3 50 1 6

2 0 6

6


2

8 1 51 6 22 4 0

X 17 3 92 11 35 7 2

X =1 20 1

2 1


2X

1 0 12 2 12 3 3

+X1 2 54 5 3

5 4 2

X = (1 2 02 3 1

)

12. Cho A =

0 1 1 11 0 1 11 1 0 1 1 1 1 0

. Tm A1.

1.4 Hng ca ma trn

1. Tm hng ca cc ma trn sau

1)

2 1 3 2 44 2 5 1 72 1 1 8 2

; 2)

1 3 5 12 1 3 45 1 1 77 7 9 1

; 3)

4 3 5 2 38 6 7 4 24 3 8 2 74 3 1 2 58 6 1 4 6

4)

1 3 1 67 1 3 1017 1 7 223 4 2 10

; 5)

0 1 10 32 0 4 5216 4 52 98 1 6 7

; 6)

2 2 1 5 11 0 4 2 12 1 5 0 11 2 2 6 13 1 8 1 11 2 3 7 2

2. Tm m hng ca ma trn A =

1 2 3 45 8 11 m+ 152 3 4 53 5 7 m+ 10

bng 2.p s : m = 1.

3. Bin lun hng ca cc ma trn sau theo tham s m

A =

3 m 1 21 4 7 21 10 17 44 1 3 3

; B =1 2 1 1 1m 1 1 1 11 m 0 1 11 2 2 1 1

; C =

3 1 1 4m 4 10 11 7 17 32 2 4 3

7

Chng 2

H PHNG TRNH TUYN TNH

1. Cho h phng trnh Ax = B 5 1 1 2 126 7 4 2 1

31 8 5 4 2

x =abc

. Tm iu kinca a, b, c h c nghim.p s : a b+ c = 0.

2. nh m h phng trnh sau v nghim

x+my + z = mx+ 2y + 2z = 12x+ (m+ 2)y + (m2 + 2)z = m2 +m

p s : m = 2.

3. Tm m 2 h sau c nghim chung{x y + z + 2t = 2m2x 3y 2z 5t = 2 v

{2x+ 3y + z 5t = 3m5x 9y 11z 26t = 1

p s : m = 32.

4. Gii cc h phng trnh sau

1)

2x y z = 43x+ 4y 2z = 113x 2y + 4z = 11

; 2)

x+ y + 2z = 12x y + 2z = 44x+ y + 4z = 2

; 3)

x 3y + 4z + t = 12x 5y + z 5t = 25x 13y + 6z = 5

4)

x+ 2y + 4z = 315x+ y + 2z = 293x y + z = 10

; 5)

x+ y + 2z + 3t = 13x y z 2t = 42x+ 3y z t = 6x+ 2y + 3z t = 4

; 6)

x+ 2y + 3z + 4t = 52x+ y + 2z + 3t = 13x+ 2y + z + 2t = 14x+ 3y + 2z + t = 5

7)

y 3z + 4t = 5x 2z + 3t = 43x+ 2y 5t = 124x+ 3y 5z = 5

; 8)

x 2y + z + t = 1x 2y + z t = 1x 2y + z + 5t = 5

; 9)

x+ y 3z = 12x+ y 2z = 1x+ y + z = 3x+ 2y 3z = 1

5. Tm m h

x+ 3y + 4z t = 22x+ 7y + 4z + t = m+ 11x+ 5y 4z + 5t = m+ 9

c nghim v gii vi m .

8

Chng 3

KHNG GIAN VECTOR

3.1 Khng gian vector

1. Trong R3 , trong cc h sau, h no l h ph thuc tuyn tnh

A A = {u1 = (5, 4, 3), u2 = (3, 3, 2), u3 = (8, 1, 3)},B B = {u1 = (2,1, 3), u2 = (3,1, 5), u3 = (1,4, 3)}C C = {u1 = (1, 2, 3), u2 = (4, 5, 6), u3 = (7, 8, 9)}D D = {u1 = (0, 1, 2), u2 = (1, 2, 7), u3 = (0, 4, 4)}.

2. Cho P2 l tp hp cc a thc bc b hn hoc bng 2 vi h s thc. Chng minh rng

a. H A = {p1(x) = 1 + 2x + 3x2, p2(x) = 2 + 3x + 4x2, p3(x) = 3 + 5x + 7x2} l phthuc tuyn tnh.

b. H B = {q1(x) = 1, q2(x) = 1 + x, q3(x) = 1 + x+ x2} l c lp tuyn tnh.c. H {p(x), p(x), p(x)}, trong p(x), p(x) l o hm cp 1 v cp 2 ca p(x) =

ax2 + bx+ c; a, b, c R l c lp tuyn tnh.3. Chng minh rng tp hp F = {y = (y1, y2, y3, y4)|y2 + y3 + y4 = 0} l mt khng gian

vector con ca R4.

4. Tm iu kin vector (x, y, z) khng phi l mt t hp tuyn tnh ca hF = {u = (1, 2, 1), v = (1, 1, 0), w = (3, 6, 3)}.p s : y = x+ z.

5. Trong R4, vi W = {u1, u2, u3} = {(1, 1, 1, 0), (0,2, 1, 1), (1, 0, 1,2)}. Chou = (x1, x2, x3, x4) R4. Tm iu kin u W .p s : 7x1 + 2x2 + 5x3 x4 = 0.

6. Trong R3 xt hai c s A, B. Bit ma trn chuyn c s t A sang B l P =

4 0 11 4 41 1 2

v ta x i vi c s A l [x]A = (13, 13, 13). Tm ta ca x i vi c s B.p s : [x]B = (1,6, 9).

7. Tm ta (x1, x2, x3, x4) ca vector u = (1, 1, 1, 1) theo c s {u1 = (0, 1, 1, 1), u2 =(1, 0, 1, 1), u3 = (1, 1, 0, 1), u4 = (1, 1, 1, 0)}.

9

8. Tm ta (x1, x2, x3) ca vector u = (m,m, 4m) theo c s {u1 = (1, 2, 3), u2 =(3, 7, 9), u3 = (5, 10, 16)}.p s : x1 = m, x2 = m, x3 = m.

9. Cho bit ma trn chuyn c s t c s U = {u1, u2, u3} sang c s chnh tc E l

A =

1 1 20 1 01 1 1

. Tm ta (x1, x2, x3) ca vector u = (1, 0, 1).p s : x1 = 3, x2 = 0, x3 = 2.

10. Trong khng gian R3 cho hai c s, c s chnh tc E vF = {f1 = (1, 1, 1); f2 = (1,1, 1); f3 = (1, 1,1)}. Hy tm ma trn chuyn c s tF sang E?

p s : PFE =

0 0.5 0.50.5 0 0.50.5 0.5 0

.11. Tm s chiu v c s ca khng gian con khng gian R3 cc nghim ca h phng

trnh thun nht

x1 2x2 + x3 = 02x1 x2 x3 = 02x1 + 4x2 2x3 = 0

12. Tm s chiu v c s ca khng gian con khng gian R4 cc nghim ca h phng

trnh thun nht

x1 + 2x2 + 3x3 + 4x4 = 01

2x1 + x2 +

3

2x3 + 2x4 = 0

1

3x1 +

2

3x2 + x3 +

4

3x4 = 0

1

4x1 +

1

2x2 +

3

4x3 + x4 = 0

13. S = {x1, x2, x3, x4, x5} l mt h vector trong R4. Tm hng ca S nu x1 =(1, 1,1,1);x2 = (1,1, 1,1);x3 = (3, 1,1, 1);x4 = (3,1, 1,1);x5 = (2, 0, 0, 0).

3.2 Khng gian Euclide

1. Trong khng gian EUCLIDE R3 vi tch v hng thng thng, cho ba vector x =(2, b, c); y = (1,2, 2); z = (2, 2, a). Tm a, b, c ba vector trn to thnh mt h trcgiao.

2. Trc giao ha v trc chun ha Gram-Schmidt h cc vector x1 = (1, 2, 3) v x2 =(3, 1, 2).

p s : y1 =

(114,

214,

314

); y2 =

(311050

,81050

,51050

).

3. Trc giao ha v trc chun ha Gram-Schmidt h cc vector x1 = (1, 1, 1); x2 =(1, 1, 0) v x2 = (1, 0, 0).

4. Trong khng gian EUCLIDE R3 cho khng gian vector conW = {x R3|2x1+x2x3 =0}. Tm mt c s trc giao v mt c s trc chun ca W .

5. Trong khng gian EUCLIDE R4 cho khng gian vector con W = {x R4|x1 +x2 +x3 =0, x1 + x2 + x4 = 0}. Tm mt c s v mt c s trc chun ca W .

10

Chng 4

NH X TUYN TNH

4.1 nh x tuyn tnh

1. Trong cc nh x sau, nh x no l nh x tuyn tnh

1. f : R3 R3, f(x1, x2, x3) = (x2 x3, x1 + x3, 3x1 x2 + 2x3)2. f : R3 R3, f(x1, x2, x3) = (x1 + x2, x2 + 2, x3 + 3)3. f : R2 R, f(x1, x2, ) = |x2 x1|4. f : R2 R2, f(x1, x2) = (2x1, x2)5. f : R2 R2, f(x1, x2) = (x21, x2)6. f : R2 R2, f(x1, x2) = (x2, x1)7. f : R2 R2, f(x1, x2) = (0, x2)8. f : R2 R2, f(x1, x2) = (x1, x2 + 1)9. f : R2 R2, f(x1, x2) = (2x1 + x2, x1 x2)10. f : R2 R2, f(x1, x2) = (x2, x2)11. f : R2 R2, f(x1, x2) = ( 3x1, 3x2)12. f : R3 R3, f(x1, x2, x3) = (x1, x1 + x3 + x2)13. f : R3 R3, f(x1, x2, x3) = (0, 0)14. f : R3 R3, f(x1, x2, x3) = (1, 1)15. f : R3 R3, f(x1, x2, x3) = (2x1 + x2, 3x2 4x3)

2. Hy tm ma trn chnh tc ca mi nh x tuyn tnh sau

1. f(x1, x2) = (2x1 x2, x1 + x2)2. f(x1, x2) = (x1, x2)

3. f(x1, x2, x3) = (x1 + 2x2 + x3, x1 + 5x2, x3)

4. f(x1, x2, x3) = (4x1, 7x2,8x3)5. f(x1, x2, ) = (x2,x1, 3x2 + x1, x1 x2)6. f(x1, x2, x3, x4) = (7x1 2x2 x3 + x4, x2 + x3,x1)7. f(x1, x2, x3) = (0, 0, 0, 0, 0)

8. f(x1, x2, x3, x4) = (x4, x1, x3, x2, x1 x3)

11

3. Cho nh x tuyn tnh f : R4 R4, nh bi

f(x, y, z, t) = (x+ 2y+ 4z 3t, 3x+ 5y+ 6z 4t, 4x+ 5y 2z+ 3t, 3x+ 8y+ 24z 19t).

Xt khng gian vector con V = {(x, y, z, t)/f(x, y, z, t) = (0, 0, 0, 0)}. Tm s chiu vmt c s ca V .p s : khng gian vector V c s chiu bng 2 v mt c s ca n{v = (8,6, 1, 0), u = (7, 5, 0, 1)}.

4. Cho T : R2 R2 l nh x nhn vi ma trn(

2 18 4

)1. Vector no sau y Im(T ): (1,-4); (5,0); (-3,12).2. Vector no sau y Ker(T ): (5,10); (3,2); (1,1).

5. Tm nhn v nh ca cc nh x tuyn tnh sau

1. f : R3 R3, f(x1, x2, x3) = (x1 2x2 + x3, 2x1 x2 x3, x1 + x2 2x3)2. f : R3 R3, f(x1, x2, x3) = (x1 + x2 + x3, x1 + x2 + x3, x1 + x2 + x3)

6. Cho f : R4 R3, v A =1 3 2 22 1 2 1

1 2 0 1

. Vi f(x) = AX, X R4, hy xc nhnhn v nh ca nh x tuyn tnh f .

7. f l mt nh x ma trn xc nh nh sau

A =

1 1 35 6 47 4 2

; B =2 0 14 0 2

0 0 0

;C =

(4 1 5 21 2 3 0

); D =

1 4 5 0 93 2 1 0 11 0 1 0 12 3 5 1 8

Hy tm1. Mt c s v s chiu cho Im(f);2. Mt c s v s chiu cho Ker(f);

8. Cho f : R2 R2 l nh x tuyn tnh c tnh cht f(1, 1) = (2, 0); f(0, 1) = (3, 1).Tnh f(1, 0) v tm ma trn ca f trong c s chnh tc ca R2.

9. Cho nh x tuyn tnh f : R2 R2, ma trn ca f i vi c s F = {(2, 1), (1, 1)} l(2 21 1

). Hy tm biu thc ca f .

p s : f(x, y) = (5y, 3y).

10. Xt c s S = {v1, v2, v3}, trong R3 trong v1 = (1, 2, 3), v2 = (2, 5, 3), v3 = (1, 0, 10).Tm cng thc biu din nh x tuyn tnh f : R3 R2 xc nh bi T (v1) =(1, 0), T (v2) = (1, 0), T (v3) = (0, 1). Tnh T (1, 1,1), trong cc c s chnh tc caR3 ,R2.

12

4.2 Gi tr ring - vector ring

1. Tm cc gi tr ring v vector ring ca cc ma trn

A =

(6 44 2

); B =

(5 22 8

); C =

(9 1212 6

)2. Tm cc gi tr ring v vector ring ca cc ma trn

A =

2 1 11 2 10 0 1

; B = 3 1 11 5 1

1 1 3

; C =6 2 22 3 4

2 4 3

3. Cho ma trn A =

8 9 910 13 104 6 3

, hy tm cc gi tr ring ca ma trn A?p s : {2, 1, 3}

4. Tm tr ring thc v vector ring ca ma trn A =

3 3 21 1 23 1 0

v xc nh cckhng gian vector ring tng ng.


2 1 00 1 10 2 4

v xc nh cc khnggian vector ring tng ng.


2 2 11 3 11 2 2

v xc nh cc khnggian vector ring tng ng.

7. Tm tr ring v vector ring ca cc ma trn sau, t hy cho ha cc ma trn (nu

c) A =

15 18 169 12 84 4 6

; B = 0 8 61 8 7

1 14 11

; C = 2 0 11 1 12 0 1

13

Chng 5

DNG TON PHNG

1. Vit ma trn ca cc dng ton phng sau:

1. f(x1, x2) = 3x21 4x1x2 x22

2. f(x1, x2, x3) = x21 2x1x2 x1x3

3. f(x1, x2, x3) = 2x21 2x22 + 5x23 8x1x2 16x1x3 + 14x2x3

4. f(x1, x2, x3) = 2x1x2 6x2x3 + 2x3x15. f(x1, x2, x3) = 2x

21 + 3x1x2 + 4x3x1 + x

22 + x

23

6. f(x1, x2, x3) = 4x1x2 4x1x3 + 3x22 2x3x2 + 3x237. f(x1, x2, x3) = x

21 + x

22 + 3x

23 + 4x1x2 + 2x1x3 + 2x2x3

8. f(x1, x2, x3) = x21 2x22 + x23 + 2x1x2 + 4x1x3 + 2x2x3

9. f(x1, x2, x3) = x21 3x23 2x1x2 + 2x1x3 6x2x3

2. a v dng chnh tc dng ton phng

1. f(x1, x2, x3) = 2x1x2 6x2x3 + 2x3x12. f(x1, x2, x3) = 2x

21 + 3x1x2 + 4x3x1 + x

22 + x

23

3. f(x1, x2, x3) = 4x1x2 4x1x3 + 3x22 2x3x2 + 3x234. f(x1, x2, x3) = x

21 + x

22 + 3x

23 + 4x1x2 + 2x1x3 + 2x2x3

5. f(x1, x2, x3) = x21 2x22 + x23 + 2x1x2 + 4x1x3 + 2x2x3

6. f(x1, x2, x3) = x21 3x23 2x1x2 + 2x1x3 6x2x3

3. Cho dng ton phng Q(x) = x21 + 2x22 + 2x

23 + 2x1x2 2x1x3 = xTAx. Bng php bin

i trc giao, v vi c s trc chun{y1 =

(26,1

6,

16

), y2 =

(13,

13,1

3

), y3 =

(0,

12,

12

)}.

Hy a dng ton phng ny v dng chnh tc.p s : g(z) = 3z22 + 2z

23

4. Kho st tnh cht xc nh (du) ca dng ton phng sauf(x1, x2, x3) = 5x

21 + x

22 + 4x1x3 4x3x2 + 5x23

5. Kho st tnh cht xc nh (du) ca dng ton phng sauf(x1, x2, x3) = 3x

21 + x

22 + 5x

23 + 4x1x2 8x1x3 4x2x3

6. nh m dng ton phng sau xc nh mf(x1, x2, x3) = 5x21 x22 mx23 4x1x2 + 2x1x3 + x2x3

14

Ti liu tham kho

[1] Trn Lu Cng (Ch bin), Nguyn nh Huy, Hunh B Ln, Nguyn B Thi, NguynQuc Ln, Ton Cao Cp 2 i S Tuyn Tnh, Nh xut bn gio dc, 2005.

[2] Nguyn nh Tr (Ch bin), L Trng Vinh, Dng Thy V, Bi tp Ton Hc CaoCp Tp 1 (Dng cho sinh vin cc trng cao ng). Nh xut bn gio dc Vit Nam,2010.

[3] Nguyn nh Tr (Ch bin), T Vn nh, Nguyn H Qunh. Bi tp TON CAO

CP Tp mt i s v hnh hc gii tch. Nh xut bn gio dc, 2010.

15

tai lieu bai tap toan cao cap a2 c2 docx

Documents