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CAÙC DAÏNG BAØI TAÄP CAÀN KHAI THAÙC
A) . DAÏNG 1: Phaân tích ña thöùc thaønh nhaân töû baèng phöông phaùp ñaët nhaân töû chung:
+ Baøi taäp :
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû
a) 3x – 3y
b) 2x2 + 5x3 + x2y
c) 14x2y – 21 xy2 + 28x2y2
d) x(y – 1 ) – y(y – 1)
e) 10x(x – y) – 8y(y – x)
Giaûi:
a) 3x – 3y = 3(x – y)
b) 2x2 + 5x3 + x2y = x2(2 + 5x + y)
c) 14x2y – 21 xy2 + 28x2y2 = 7xy( 2x – 3y + 4xy)
d) x(y – 1 ) – y(y – 1) = (y – 1)(x – y)
e) 10x(x – y) – 8y(y – x) = 10x(x – y) + 8y(x – y) = 2 (x – y)(5x + 4y)
2) Tìm x , bieát :
a) 5x(x – 2000) – x + 2000 = 0
b) 5x2 = 13x
Giaûi:
a) Ta coù : 5x(x – 2000) – x + 2000 = 0
5x(x – 2000) – (x – 2000) = 0
(x – 2000)(5x – 1) = 0
x – 2000 = 0 hoaëc 5x – 1 = 0
· x – 2000 = 0 x = 2000
· 5x – 1 = 0 5x = 1 x =
Vaäy x = 2000 hoaëc x =
b) 5x2 = 13x 5x2 – 13x = 0
x(5x – 13 ) = 0
5x = 0 hoaëc 5x – 13 = 0
· x = 0
· 5x – 13 = 0 x =
Vaäy x = 0 hoaëc x =
3) Chöùng minh raèng : 55n+1 – 552 chia heát cho 54 ( Vôùi n laø soá töï nhieân )
Giaûi:
Ta coù : 55n+1 – 55 = 55n.55 – 55n
= 55n(55 – 1) = 55n.54
Maø 54 chia heát cho 54 neân 55n.54 ( ñpcm)
4 ) Tính nhanh
a) 15,8 . 35 + 15,8 . 65
b) 1,43 . 141 – 1.43 . 41
Giaûi:
a) 15,8 . 35 + 15,8 . 65 = 15,8(35 + 65) = 15,8 . 100 = 1580
b) 1,43 . 141 – 1.43 . 41 = 1,43 ( 141 – 41 ) 1,43 . 100 =143
+ Baøi taäp töông töï:
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû
a) 6x4 – 9x3
b) x2y2z + xy2z2 + x2yz2
c) (x + y ) 3 – x3 – y3
d) 2x(x + 3) + 2(x + 3)
2) Tìm x , bieát
a) 5x(x – 2) – x – 2 = 0
b) 4x(x + 1) = 8( x + 1)
c) x(2x + 1) + = 0
d) x(x – 4) + (x – 4)2 = 0
3) Chöùng minh raèng :
a) Bình phöông cuûa moät soá leû chia cho 4 thì dö 1
b) Bình phöông cuûa moät soá leû chia cho 8thì dö 1
+ Khaùi quat hoùa baøi toaùn :
Phaân tích ña thöùc sau thaønh nhaân töû:
A = pm+2.q – pm+1.q3 – p2.qn+1+ p.qn+3
+ Ñeà xuaát baøi taäp töông töï:
Phaân tích caùc ña thöùc sau thaønh nhaân töû:
a) 4x(x – 2y) + 8y(2y – x )
b) 3x(x + 7)2 – 11x2(x + 7 + 9(x + 7)
c) -16a4b6 – 24a5b5 – 9a6b4
d) 8m3 + 36m2n + 54mn2 + 27n3
B) . DAÏNG 2: Phaân tích ña thöùc thaønh nhaân töû baèng phöông phaùp dung haèng ñaúng thöùc
+ Baøi taäp :
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû :
a) x2 + 6x + 9
b) 10x – 25 – x2
c) (a + b)3 + (a – b)3
d) (a + b)3 – (a – b)3
e) x3 + 27
f) 81x2 – 64y2
g) 8x3 + 12x2y + 6xy2 + y3
Giaûi:
a) x2 + 6x + 9 = x2+ 2 .x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -( x2 – 2.x.5 + 52) = - (x – 5)2
c) (a + b)3 + (a – b)3= [(a + b) + (a – b)][(a + b)2 – (a + b)(a – b) + (a – b)2
= 2a[a2 + 2ab + b2 – (a2- b2) + a2 – 2ab + b2
= 2a(a2 + 3b2)
d) (a + b)3 – (a – b)3 = [(a + b) - (a – b)][(a + b)2 + (a + b)(a – b) + (a – b)2]
= ( a + b – a + b) (a2 + 2ab + b2 + a2- b2+ a2 – 2ab + b2
= 2b(3a2+ b2)
e) x3 + 27 = ( x + 3)(x2 – 3x + 9)
f) 81x2 – 64y2 = (9x)2 – (8y)2 = (9x + 8y)(9x – 8y)
g) 8x3 + 12x2y + 6xy2 + y3 = (2x)3 + 3.(2x)2.y + 3.(2x).y2 + y3
= (2x + y)3
2) Tìm x , bieát :
a) x2 – 25 = 0
b) x2 – 4x + 4 = 0
Giaûi :
a) x2 – 25 = 0
( x – 5 )(x + 5) = 0
b) x2 – 4x + 4 = 0 x2 – 2.2x + 22 = 0
(x – 2)2 = 0
x – 2 = 0
x = 2
3) Chöùng minh raèng hieäu caùc bình phöông cuûa hai soá leû lieân tieáp thì chia heát cho 8
Giaûi:
Goïi hai soá leû lieân tieáp laø 2a – 1 vaø 2a + 1 ( a laø soá nguyeân ) . Hieäu caùc bình phöông cuûa chuùng laø: ( 2a + 1)2 – (2a – 1)2.
Ta thaáy ( 2a + 1)2 – (2a – 1)2. = (2a + 1 + 2a – 1 )(2a + 1 -2a + 1)
= 4a.2 = 8a chia heát cho 8
4)Tính nhaåm:
c) 732 – 272
d) 372 – 132
e) 20022 – 22
Giaûi:
a) 732 – 272 = ( 73 + 27) (73 – 27) = 100 . 46 = 4600
b) 372 – 132 = (37 – 13 )(37 + 13) = 24 . 50 = 1200
c) 20022 – 22 = (2002 – 2)(2002 + 2) = 2000 . 2004 = 4008000
+ Baøi taäp töông töï:
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû:
a) ( a + b + c)3 – a3 – b3 – c3
b) 8(x + y + z)3 – (x + y)3 – (y + z)3 – (z – x)3
c) 8x3 – 27
d) – x3 + 9x2 – 27x + 27
2) Tìm x , bieát :
a) 4x2 – 49 = 0
b) x2 + 36 = 0
3) Chöùng minh raèng vôùi moïi soá nguyeân n ta coù : (4n + 3)2 – 25 chia heát cho 8
4) Tính nhanh giaù trò cuûa bieåu thöùc sau vôùi a = 1982
M = (a + 4)2 + 2(a + 4)(6 – a) + (6 – a)2
+ Khaùi quat hoùa baøi toaùn :
- Chöùng minh hieäu caùc bình phöông cuûa hai soá leû lieân tieáp thì chia heát cho 8
- Chöùng minh hieäu caùc bình phöông cuûa hai soá chaúnû lieân tieáp thì chia heát cho 16
+ Ñeà xuaát baøi taäp töông töï:
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû
a) ( 3x – 2y)2 – (2x + y)2
b) 27x3 – 0,001
c) [4abcd + (a2 + b2)(c2 + d2)]2 – 4[cd(a2 + b2) + ab(c2 + d2)]2
d) x6 + 2x5 + x4 – 2x3 – 2x2 + 1
2) Chöùng minh raèng bieåu thöùc : 4x(x + y) ( x + y + z)(x + y) y2z2 luoân luoân
khoâng aâm vôùi moïi giaù trò cuûa x , y vaø z
C) . DAÏNG 3: Phaân tích ña thöùc thaønh nhaân töû baèng phöông phaùp nhoùm haïng töû
+ Baøi taäp :
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû :
a) x2 + 4x – y2 + 4
b) 3x2 + 6xy + + 3y2 – 3z2
c) x2 – 2xy + y2 – z2 + 2zt - t2
d) x2(y – z) + y2(z – x) + z2(x – y)
Giaûi:
a) x2 + 4x – y2 + 4 = x2 +2.x.2 + 22 – y2
= (x + 2)2 – y2 = (x + 2 – y)(x + 2 + y)
b) 3x2 + 6xy + + 3y2 – 3z2 = 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2] = 3(x + y + t)(x + y – z)
c) x2 – 2xy + y2 – z2 + 2zt - t2 = (x2 – 2xy + y2) – (z2 - 2zt + t2)
= (x – y)2 – (z – t)2 = (x – y + z – t )(x – y – z + t)
d) x2(y – z) + y2(z – x) + z2(x – y)
+ Caùch 1: Khai trieån hai soá haïng cuoái roài nhoùm caùc soá haïng laøm xuaát hieän nhaân töû
chung y – z
x2(y – z) + y2(z – x) + z2(x – y) = x2(y – z) + y2z – y2x + z2x – z2y
= x2(y – z) + yz(y – z) – x(y2- z2)
= (y – z)(x2 + yz – xy – xz)
= (y – z)[x(x – y) – z(x – y)]
= (y – z )(x – y)(x – z)
+ Caùch 2:Taùch z – x = -[(y – z) + (x –y)]
x2(y – z) + y2(z – x) + z2(x – y) = x2(y – z) – y2[(y – x) + (x – y)] + z2(x – y)
= (y – z)(x2 - y2) – (x – y)(y2 – z2)
= (y – z)(x + y)(x – y) – (x – y)(y + z)(y – z)
= (y – z)(x – y)(x + y – y – z )
= (y – z)(x – y)(x – z)
2) Tìm x , bieát :
a) x(x – 2) + x – 2 = 0
b) 5x(x – 3) – x + 3 = 0
Giaûi:
a) x(x – 2) + x – 2 = 0 (x – 2)(x + 1) = 0
x – 2 = 0 hoaëc x +1 = 0
x = 2 hoaëc x = -1
b) 5x(x – 3) – x + 3 = 0 5x(x – 3) – (x – 3) = 0
(x – 3)(5x – 1) = 0
x – 3 = 0 hoaëc x – 1 = 0
x = 3 hoaëc x = 1
+ Baøi taäp töông töï:
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû :
a) x3 + 3x2y + x + 3xy2 + y + y3
b) x3 + y(1 – 3x2) + x(3y2 – 1) – y3
c) 27x3 + 27x2 + 9x + 1 + +
d) x2y + xy2 – x – y
e) 8xy3 – 5xyz – 24y2 + 15z
2) Tìm x , bieát :
a) x2 – 6x + 8 = 0
b) 9x2 + 6x – 8 = 0
c) x3 + x2  + x + 1 = 0
d) x3 - x2 - x + 1 = 0
+ Khaùi quaùt hoùa baøi toaùn :
Phaân tích ña thöùc thaønh nhaân töû : pm + 2 q – pm + 1 q3 – p2 qn + 1 + pq n + 3
+ Ñeà xuaát baøi taäp:
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû:
a) bc(b + c) + ac(c – a) – ab(a + b)
b) x(x + 1)2 + x(x – 5) – 5(x + 1)2
c) ab(a – b) + bc(b – c) + ca(c – a)
d) x3z + x2yz – x2z2 – xyz2
2) Tìm taát caû caùc giaù trò cuûa x , y sao cho: xy + 1 = x + y
3) Phaân tích ña thöùc thaønh nhaân töû roài tính giaù trò cuûa ña thöùc vôùi x = 5,1 ; y = 3,1 cuûa ña thöùc : x2 – xy – 3x + 3y
D) . DAÏNG 4: Phaân tích ña thöùc thaønh nhaân töû baèng caùch phoái hôïp nhieàu phöông phaùp
+ Baøi taäp :
1) Phaân tích caùc ña thöùc sau thaønh nhaân töû:
a) a3 + b3 + c3 – 3abc
b) (x – y )3 + (y – z )3 + (z – x)3
Giaûi:
a) •° Caùch 1:
a3 + b3 + c3 – 3abc = (a + b)3 – 3ab(a + b) + c3 – 3abc
= (a + b)3 + c3 – 3ab(a + b) – 3abc
= (a + b + c)[(a + b)2 – (a + b) c + c2] – 3ab(a + b + c)
= (a + b + c)(a2 + 2ab + b2 – ac –bc + c2 – 3ab
=(a + b + c)(a2 + b2 + c2 – ab – bc – ca )
• ° Caùch 2:
a3 + b3 + c3 – 3abc = a3 + a2b + a2c + b3 + ab2 + b2c + c3 + ac2 + bc2 – a2b – abc - a2c – ac2 – abc –b2c – abc – bc2
= a2(a + b + c) + b2(b + a + c) + c2(c + a + b) – ab(a + b + c)
– ac((a + c + b) – bc(b + a + c)
= (a + b + c)(a2 + b2 + c2 – ab – ac – bc)
b) • ° Caùch 1:
Ñaët x – y = a ; y – z = b ; z – x = c, thì a + b + c = 0
Khi ñoù theo caâu a ta coù : a3 + b3 + c3 – 3abc = 0
Hay a3 + b3 + c3 = 3abc
Vaäy (x – y )3 + (y – z )3 + (z – x)3 = 3(x – y)(y – z)(z – x)
•° Caùch 2:
Ñeå yù raèng (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + 3ab(a + b) + b3
Vaø (y – z) = (y – x) + (x – z )
Do ñoù : (x – y)3 + (y –z )3 + (z – x)3 = [(y – x) + (x – z)]3 + (z – x)3 + (x – y)3
= (y – x)3 +3(y – x)(x –z)[( y – x) + (x –z)]+
+ (x – z)3 – (x –z )3 – (y – x)3
= 3(x – y)(y – z)(z – x)
•° Caùch 3: Khai trieån caùc haèng ñaúng thöùc roài söû duïng phöông phaùp ñaët thöøa soá
chung
(x – y )3 + (y – z )3 + (z – x)3 = x3 – 3x2y + 3xy2 – y3 + y3 – 3y2z + 3yz2 – z3 + z3
– 3z2x + 3zx2 – x3
= - 3x2y + 3xy2 – 3y2z + 3yz2 – 3z2x + 3zx2
= 3(-x2y + xy2 – y2z + yz2 – z2x + zx2)
= 3[-xy(x – y) – z2(x – y) + z(x – y)(x + y)]
= 3(x – y)( - xy – z2 + xz + yz)
= 3(x – y)[y(z – x) – z(z – x)]
= 3(x – y)(z – x)(y –z )
2) Phaân tích ña thöùc sau thaønh nhaân töû baèng phöông phaùp taùch caùc haïng töû:
x3 – 7x – 6
Giaûi:
° Caùch 1: Taùch soá haïng -7x thaønh –x – 6x , ta coù :
x3 – 7x – 6 = x3 – x – 6x – 6
= (x3 – x) – (6x + 6)
= x(x + 1)(x – 1) – 6(x + 1)
= (x + 1)(x2 – x – 6)
Ñeå tieáp tuïc phaân tích ña thöùc x2 – x – 6 thaønh nhaân töû , ta laïi taùch soá haïng – 6 thaønh – 2 – 4 . Khi ñoù :
x3 – 7x – 6 = (x + 1)(x2 – x – 2 – 4 )
= (x + 1)[(x + 2)(x – 2) – (x + 2)]
= (x + 1)(x + 2)(x – 3)
° Caùch 2 : Taùch soá haïng – 7x thaønh – 4x – 3x , ta coù:
x3 – 7x – 6 = x3 – 4x – 3x – 6
= x( x + 2)(x – 2) – 3(x + 2)
= (x + 2)(x2 – 2x – 3)
Tieáp tuïc taùch soá haïng – 3 cuûa nhaân töû thöù hai thaønh – 1 – 2 , Ta coù :
x3 – 7x – 6 =(x + 2)(x2 – 1 – 2x – 2)
= (x + 2)[(x – 1)(x + 1) – 2( x + 1)]
= (x + 2)(x + 1)(x – 3 )
° Caùch 3: Taùch soá haïng – 6 = 8 – 14 , Ta coù:
x3 – 7x – 6 = x3 + 8 – 7x – 14
= (x + 2)(x2 – 2x + 4) – 7(x + 2)
= (x + 2)(x2 – 2x – 3)
Tieáp tuïc taùch soá haïng – 3 thaønh + 1 – 4 , Ta coù :
x3 – 7x – 6 = (x + 2)(x2 – 2x + 1 – 4 )
= (x + 2)[(x – 1)2 – 22]
= (x + 2)(x + 1)(x – 3)
3) Duøng phöông phaùp ñaët aån phuï , phaân tích ña thöùc thaønh nhaân töû:
a) (x2 + x + 1)(x2 + x + 2) – 12
b) 4x(x + y)(x + y + z )(x + z) + y2z2
Giaûi:
Ñaët: x2 + x + 1 = y , ta coù x2 + x + 2 = y + 1 . Ta coù:
(x2 + x + 1)(x2 + x + 2) – 12 = y(y + 1) – 12
= y2 + y – 12
= y2 – 9 + y – 3 = (y – 3)(y + 3) + (y – 3)
= (y – 3)(y + 4)
Thay x2 + x + 1 = y , ta ñöôïc :
(x2 + x + 1 – 3)( x2 + x + 1 + 4) = (x2 + x – 2)( x2 + x + 5)
= [(x – 1)(x + 1) + (x – 1)]( x2 + x + 5)
= (x - 1)(x + 2)( x2 + x + 5)
b)4x(x + y)(x + y + z )(x + z) + y2z2
= 4x(x + y + z)(x + y)(x + z) + y2z2
= 4(x2 + xy + xz)(x2 + xy + xz + yz) + y2z2
Ñaët : x2 + xy + xz = m , ta coù :
4x(x + y + z)(x + y)(x + z) + y2z2 = 4m(m + yz) + y2z2
= 4m2 + 4myz + y2z2 = (2m + yz)2
Thay m = x2 + xy + xz , ta ñöôïc :
(x + y)(x + y + z )(x + z) + y2z2 = (2x2 + 2xy + 2xz + yz)2
4) Duøng phöông phaùp heä soá baát ñònh ñeå :
a) Phaân tích ña thöùc x3 – 19x – 30 thaønh tích hai ña thöùc baäc nhaát vaø baäc hai
b) Phaân tích ña thöùc x4 + 6x3 + 7x2 + 6x + 1
Giaûi:
a) Keát quaû caàn phaûi tìm coù daïng :
(x + a)(x2 + bx + c) = x3 + (a + b)x2 + (ab + c)x + ac
Ta phaûi tìm boä soá a , b , c thoûa maõn:
x3 – 19x – 30 = x3 + (a + b)x2 + (ab + c)x + ac
Vì hai ña thöùc naøy ñoàng nhaát , neân ta coù:
Vì a , c Z vaø tích ac = - 30 , do ñoù a , c
Vaø a = 2 , c = -15 , Khi ñoù b = -2 thoûa maõn heä thöùc treân . Ñoù laø boä soá phaûi
tìm , töùc laø : x3 – 19x – 30 = (x + 2)(x2 – 2x – 15)
b) Deå thaáy raèng 1 khoâng laø nghieäm cuûa ña thöùc neân ña thöùc khoâng coù nghieäm nguyeân , cuõng khoâng coù nghieäm höõu tæ .
Nhö vaäy neáu ña thöùc ñaõ cho phaân tích ñöôïc thaønh thöøa soá thì phaûi coù daïng
(x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd
Suy ra :
Töø heä naøy ta tìm ñöôïc a = b = d = 1 , c = 5
Vaäy x4 + 6x3 + 7x2 + 6x + 1 = ( x2 + x + 1)(x2 + 5x + 1)

5) Phaân tích ña thöùc sau thaønh nhaân töû: x5 + x + 1
Giaûi:
° Caùch 1
x5 + x + 1 = x5 + x4 + x3 – x4 – x3 – x2 + x2 + x + 1
= x3(x2 + x + 1) – x2(x2 + x + 1) + 1(x2 + x + 1)
= (x2 + x + 1)(x3 – x2 + 1)
° Caùch 2 :
x5 + x + 1 = x5 – x2 + x2 + x + 1
= x2(x3 – 1) + 1(x2 + x + 1)
= x2(x – 1)(x2 + x + 1) + 1(x2 + x + 1)
= (x2 + x + 1)[(x2(x – 1) + 1]
= (x2 + x + 1)[x3 – x2 + 1)
6)Phaân tích ña thöùc sau thaønh nhaân töû : x2 – 8x + 12
Giaûi:
° Caùch 1: x2 – 8x + 12 = x2 – 2x – 6x + 12
= (x2 – 2x) – (6x – 12)
= x(x – 2) – 6(x – 2)
= (x – 2)(x – 6)
° Caùch 2 : x2 – 8x + 12 = (x2 – 8x + 16) – 4
= (x – 4)2 - 22
= (x – 4 + 2)(x – 4 – 2 )
= (x – 2 )(x – 6)
° Caùch 3 : x2 – 8x + 12 = x2 – 36 – 8x + 48
= (x2 – 36) – (8x – 48)
= (x + 6)(x – 6) – 8(x – 6)
= (x – 6)(x + 6 – 8)
= (x – 6)(x – 2)
° Caùch 4 : x2 – 8x + 12 = x2 – 4 – 8x + 16
= (x2 – 4) – (8x – 16)
= (x + 2)(x – 2) – 8(x – 2)
= (x – 2)(x + 2 – 8)
= (x – 2)(x – 6)
° Caùch 5: x2 – 8x + 12 = x2 – 4x + 4 – 4x + 8
= (x2 – 4x + 4) – (4x – 8)
= (x – 2)2 – 4(x – 2)
= (x – 2)(x – 2 – 4)
= (x – 2)(x – 6)
° Caùch 6: x2 – 8x + 12 = x2 – 12x + 36 + 4x – 24
= (x2 – 12x + 36) + (4x – 24)
= (x – 6)2 + 4(x – 6)
= (x – 6)(x – 6 + 4)
= (x – 6)(x – 2)
7)Phaân tích ña thöùc sau thaønh nhaân töû : x2 + 4xy + 3y2
Giaûi:
° Caùch 1: x2 + 4xy + 3y2 = x2 + xy + 3xy + + 3y2
= (x2 + xy) + (3xy + + 3y2)
= x(x + y) + 3y(x + y)
= (x + y)(x + 3y)
° Caùch 2 : x2 + 4xy + 3y2 = x2 + 4xy + 4y2 – y2
= (x2 + 4xy + 4y2) – y2
= (x + 2y)2 – y2
= (x + 2y + y)(x + 2y – y)
= (x + 3y)(x + y)
° Caùch 3 : x2 + 4xy + 3y2 = x2 – y2 + 4xy + 4y2
= (x2 – y2) + ( 4xy + 4y2)
= (x + y)(x – y) + 4y(x + y)
= (x + y)(x – y + 4y)
= (x + y)(x + 3y)
° Caùch 4 : x2 + 4xy + 3y2 = x2 – 9y2 + 4xy + 12y2
= (x2 – 9y2) + (4xy + 12y2)
= (x + 3y)(x – 3y) + 4y(x + 3y)
= (x + 3y)(x – 3y + 4y)
= (x + 3y)(x + y)
° Caùch 5 : x2 + 4xy + 3y2 = x2 + 2xy + y2 + 2xy + 2y2
= (x2 + 2xy + y2) + (2xy + 2y2)
= (x + y)2 + 2y(x + y)
= (x + y)(x + y + 2y)
= (x + y)( x + 3y)
° Caùch 6 : x2 + 4xy + 3y2 = x2 + 6xy + 9y2 – 2xy – 6y2
= (x2 + 6xy + 9y2) – (2xy + 6y2)
= (x + 3y)2 – 2y(x + 3y)
= (x + 3y)(x + 3y – 2y)
= (x + 3y)(x + y)
° Caùch 7 : x2 + 4xy + 3y2 = 4x2 + 4xy – 3x2 + 3y2
= (4x2 + 4xy) – (3x2 – 3y2)
= 4x(x + y) – 3(x + y)(x – y)
= (x + y)(4x – 3x + 3y)
= (x + y)(x + 3y)
8)Phaân tích ña thöùc sau thaønh nhaân töû: a3(b2 – c2) + b3(c2 – a2) + c3(a2 – b2)
Giaûi:
° Caùch 1: a3(b2 – c2) + b3(c2 – a2) + c3(a2 – b2)
= a3(b2 – c2) + b3[(c2 – b2) – (a2 – b2) ] + c3(a2 – b2)
= a3(b2 – c2) + b3(c2 – b2) – b3(a2 – b2) + c3(a2 – b2)
= (b2 – c2)(a3 – b3) – (a2 – b2)(b3 – c3)
= (b + c)(b – c)(a – b)(a2 + ab + b2) – (a + b)(a – b)(b – c)(b2 + bc + c2)
= (a – b)(b – c)[(b + c)(a2 + ab + b2) – (a + b)( b2 + bc + c2)]
= (a – b)(b – c)(a2b + ab2 + b3 + a2c + abc + b2c – ab2 – abc – ac2 – b3 – b2c – bc2
= (a – b)(b – c)(a2b + a2c – bc2 – ac2)
= (a – b)(b – c)[b(a2 – c2) + ac(a – c)]
= (a – b)(b – c)[b(a – c)(a + c) + ac(a – c)]
= (a – b)(b – c)(a – c)(ab + bc + ac)
° Caùch 2 : M = a3(b2 – c2) + b3(c2 – a2) + c3(a2 – b2)
Xem M laø ña thöùc bieán a , khi a = b thì M = 0 neân M chia heát cho a – b . Do vai troø cuûa
a , b , c gioáng nhau khi ta hoaùn vò voøng quanh neân M chia heát cho b – c , M chia heát cho c – a
Ta coù : M = (a – b)(b – c)(c – a)(ab + bc + ca). P
Cho a = - 1 , b = -1 , c = 0 ta coù P = -1
Do ñoù : a3(b2 – c2) + b3(c2 – a2) + c3(a2 – b2) = (a – b)(b – c)(a – c)(ab + bc + ca)
9)Tìm x , bieát :
a) (2x – 1)2 – (x +3)2 = 0
b) 5x(x – 3) + 3 – x = 0
Giaûi:
a) (2x – 1)2 – (x +3)2 = 0 [(2x – 1) + (x +3)][ (2x – 1) - (x +3) = 0
( 2x – 1 + x +3)( 2x – 1 – x – 3 ) = 0
(3x + 2)(x – 4 ) = 0

c) 5x(x – 3) + 3 – x = 0 5x(x – 3) – (x – 3) = 0
(x – 3)(5x – 1) = 0

10)Tìm x , bieát :
d) (5 – 2x)(2x + 7) = 4x2 – 25
e) x3 + 27 + (x + 3)(x – 9) = 0
f) 4(2x + 7) – 9(x + 3)2 = 0
g) (5x2 + 3x – 2 )2 = (4x2 – 3x – 2 )2
Giaûi
a) (5 – 2x)(2x + 7) – 4x2 + 25 = 0
(5 – 2x)(2x + 7) – (5 – 2x)(5 + 2x) = 0
(5 – 2x)( 2x + 7 – 5 – 2x ) = 0
(5 – 2x).2 = 0
5 – 2x = 0
x =
b) x3 + 27 + (x + 3)(x – 9) = 0
(x + 3)(x2 – 3x + 9 ) + ( x + 3)(x – 9) = 0
(x + 3)( x2 – 3x + 9 + x – 9) = 0
(x + 3)(x2 – 2x) = 0
x(x – 2)(x + 3) = 0

c) 4(2x + 7)2 – 9(x + 3)2 = 0
[2(2x + 7)]2 – [3(x + 3)]2 = 0
(4x + 14)2 – (3x + 9)2 = 0
(4x + 14 + 3x + 9)(4x + 14 – 3x – 9 ) = 0
(7x + 23)(x + 5) = 0

d) (5x2 + 3x – 2 )2 = (4x2 – 3x – 2 )2
(5x2 + 3x – 2 )2 - (4x2 – 3x – 2 )2 = 0
(5x2 + 3x – 2 + 4x2 – 3x – 2)( 5x2 + 3x – 2 – 4x2 + 3x + 2) = 0
(9x2 – 4 )(x2 + 6x) = 0
(3x – 2 )(3x + 2)x(x + 6) = 0

11)Chöùng minhraèng: n3 – n chia heát cho 6 vôùi moïi n Z
Giaûi:
Ta coù : n3 – n = n(n2 – 1) = n(n – 1)(n + 1)
° Vôùi moïi n Z , khi chia n cho 2 xaûy ra hai tröôøng hôïp :
+ Tröông hôïp 1: n chia heát cho 2 , khi ñoù tích n(n – 1)(n + 1) chia heát cho 2
+ Tröông hôïp2: n chia heát cho 2 dö 1 , khi ñoù n – 1 chia heát cho 2 neân tích
n(n – 1)(n + 1) chia heát cho 2

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