Đề cương ôn tập môn Toán Lớp 6 - Phân tích đa thức thành phân tử

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
3x – 3y
2x2 + 5x3 + x2y
14x2y – 21 xy2 + 28x2y2
x(y – 1 ) – y(y – 1)
10x(x – y) – 8y(y – x)
Giaûi:
3x – 3y = 3(x – y)
2x2 + 5x3 + x2y = x2(2 + 5x + y)
14x2y – 21 xy2 + 28x2y2 = 7xy( 2x – 3y + 4xy)
x(y – 1 ) – y(y – 1) = (y – 1)(x – y)
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 = 
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:
15,8 . 35 + 15,8 . 65 = 15,8(35 + 65) = 15,8 . 100 = 1580
1,43 . 141 – 1.43 . 41 = 1,43 ( 141 – 41 ) 1,43 . 100 =143 
+ Baøi taäp töông töï: 
Phaân tích caùc ña thöùc sau thaønh nhaân töû
6x4 – 9x3
x2y2z + xy2z2 + x2yz2
(x + y ) 3 – x3 – y3
2x(x + 3) + 2(x + 3)
Tìm x , bieát 
5x(x – 2) – x – 2 = 0
4x(x + 1) = 8( x + 1)
x(2x + 1) + = 0
x(x – 4) + (x – 4)2 = 0
Chöùng minh raèng :
Bình phöông cuûa moät soá leû chia cho 4 thì dö 1
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:	 
4x(x – 2y) + 8y(2y – x )
3x(x + 7)2 – 11x2(x + 7 + 9(x + 7)
-16a4b6 – 24a5b5 – 9a6b4
8m3 + 36m2n + 54mn2 + 27n3
 B) . DAÏNG 2: Phaân tích ña thöùc thaønh nhaân töû baèng phöông phaùp dung haèng ñaúng thöùc
 	+ Baøi taäp :
Phaân tích caùc ña thöùc sau thaønh nhaân töû :
x2 + 6x + 9 
10x – 25 – x2
(a + b)3 + (a – b)3
(a + b)3 – (a – b)3
x3 + 27 
81x2 – 64y2
8x3 + 12x2y + 6xy2 + y3
 Giaûi:
x2 + 6x + 9 = x2+ 2 .x . 3 + 32 = (x + 3)2
10x – 25 – x2 = -( x2 – 2.x.5 + 52) = - (x – 5)2
(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)
(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)
x3 + 27 = ( x + 3)(x2 – 3x + 9)
81x2 – 64y2 = (9x)2 – (8y)2 = (9x + 8y)(9x – 8y)
8x3 + 12x2y + 6xy2 + y3 = (2x)3 + 3.(2x)2.y + 3.(2x).y2 + y3
 = (2x + y)3
Tìm x , bieát :
x2 – 25 = 0
x2 – 4x + 4 = 0 
Giaûi :
x2 – 25 = 0
 ( x – 5 )(x + 5) = 0 
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:
732 – 272 
372 – 132 
20022 – 22 
Giaûi:
732 – 272 = ( 73 + 27) (73 – 27) = 100 . 46 = 4600
372 – 132 = (37 – 13 )(37 + 13) = 24 . 50 = 1200
20022 – 22 = (2002 – 2)(2002 + 2) = 2000 . 2004 = 4008000 
+ Baøi taäp töông töï:
Phaân tích caùc ña thöùc sau thaønh nhaân töû:
( a + b + c)3 – a3 – b3 – c3
8(x + y + z)3 – (x + y)3 – (y + z)3 – (z – x)3
8x3 – 27 
– x3 + 9x2 – 27x + 27
Tìm x , bieát :
4x2 – 49 = 0
x2 + 36 = 0
Chöùng minh raèng vôùi moïi soá nguyeân n ta coù : (4n + 3)2 – 25 chia heát cho 8
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:
Phaân tích caùc ña thöùc sau thaønh nhaân töû 
( 3x – 2y)2 – (2x + y)2
27x3 – 0,001 
[4abcd + (a2 + b2)(c2 + d2)]2 – 4[cd(a2 + b2) + ab(c2 + d2)]2
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 :
Phaân tích caùc ña thöùc sau thaønh nhaân töû :
x2 + 4x – y2 + 4
3x2 + 6xy + + 3y2 – 3z2
x2 – 2xy + y2 – z2 + 2zt - t2 
x2(y – z) + y2(z – x) + z2(x – y)
Giaûi:
x2 + 4x – y2 + 4 = x2 +2.x.2 + 22 – y2
 = (x + 2)2 – y2 = (x + 2 – y)(x + 2 + y)
3x2 + 6xy + + 3y2 – 3z2 = 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2] = 3(x + y + t)(x + y – z)
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)
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) 
Tìm x , bieát :
x(x – 2) + x – 2 = 0
5x(x – 3) – x + 3 = 0
Giaûi:
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:
Phaân tích caùc ña thöùc sau thaønh nhaân töû :
x3 + 3x2y + x + 3xy2 + y + y3
x3 + y(1 – 3x2) + x(3y2 – 1) – y3
27x3 + 27x2 + 9x + 1 + + 
x2y + xy2 – x – y 
8xy3 – 5xyz – 24y2 + 15z
Tìm x , bieát : 
x2 – 6x + 8 = 0
9x2 + 6x – 8 = 0
x3 + x2 + x + 1 = 0
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:
Phaân tích caùc ña thöùc sau thaønh nhaân töû:
bc(b + c) + ac(c – a) – ab(a + b)
x(x + 1)2 + x(x – 5) – 5(x + 1)2
ab(a – b) + bc(b – c) + ca(c – a)
x3z + x2yz – x2z2 – xyz2
Tìm taát caû caùc giaù trò cuûa x , y sao cho: xy + 1 = x + y 
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 :
Phaân tích caùc ña thöùc sau thaønh nhaân töû:
a3 + b3 + c3 – 3abc
(x – y )3 + (y – z )3 + (z – x)3
Giaûi:
 •° 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)
• ° 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 )
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)
Duøng phöông phaùp ñaët aån phuï , phaân tích ña thöùc thaønh nhaân töû:
(x2 + x + 1)(x2 + x + 2) – 12 
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:
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)
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 :
(2x – 1)2 – (x +3)2 = 0
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 
5x(x – 3) + 3 – x = 0 5x(x – 3) – (x – 3) = 0
 	 (x – 3)(5x – 1) = 0 
10)Tìm x , bieát :
(5 – 2x)(2x + 7) = 4x2 – 25
x3 + 27 + (x + 3)(x – 9) = 0
4(2x + 7) – 9(x + 3)2 = 0
(5x2 + 3x – 2 )2 = (4x2 – 3x – 2 )2 
Giaûi
(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 	= 	
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
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
(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
 ° Vôùi moïi n Z , khi chia n cho 3 xaûy ra ba tröôøng hôïp:
	+ Tröông hôïp 1: n chia heát cho 3 , khi ñoù tích n(n – 1)(n + 1) chia heát cho 3
	+ Tröôøng hôïp 2 : n chia cho 3 dö 1 , khi ñoù n – 1 chia heát cho 3 neân tích chia
 heát cho 3 
	+ Tröôøng hôïp 3: n chia cho 3 dö 2 , khi ñoù n + 1 chia heát cho 3 neân tích chia 
 heát cho 3 
	Vaäy trong moïi tröôøng hôïp n3 – n chia heát cho 2 vaø 3 .
	Do 2 vaø 3 laø hai soá nguyeân toá cuøng nhau .
	Suy ra : n3 – n chia heát cho 2 x 3 = 6 
 12) Cho a, b , c thoûa maõn a + b + c = 0 . Chöùng minh raèng : a3 + b3 + c3 = 3abc 
Giaûi:
	° Caùch 1 : 
	a + b + c = 0 a + b = - c (a + b)3 = (- c)3
	 a3 + b3 + 3ab(a + b) = - c3 a3 + b3 + 3ab(- c) = - c3
	 a3 + b3 + c3 = 3abc
	 ° Caùch 2 :
	a + b + c = 0 a + b = - c - ab(a + b) = abc 
	 - a2b – ab2 = abc 
	Töông töï: - b2c – bc2 = abc ; - c2a – ca2 = abc 
	Do ñoù : 3abc = - a2b – ab2 – b2c – bc2 – c2a – ca2
	 3abc = - a2(b + c) – b2(a + c) – c2(a + b)
	 3abc = - a2(-a) – b2(-b) – c2(-c)
	 a3 + b3 + c3 = 3abc 
	 ° Caùch 3 :
	a + b + c = 0 a + b = - c - c2(a + b) = c3
	 -a2c – bc2 = c3
	Töông töï : -ab2 – cb2 = b3 ; -ba2 – ca2 = a3
	Do ñoù : -ab2 – cb2 – ab2 – cb2 – ba2 – ca2 = a3 + b3 + c3
	 - ac( c + a) – bc(c + b) – ab(b + a) = a3 + b3 + c3
	 -ac(-b) – bc(-a) – ab(-c) = a3 + b3 + c3
	 a3 + b3 + c3 = 3abc
13)Tính nhanh : 
x2 + vôi x = 49,75
x2 – y2 – 2y – 1 vôùi x = 93 , y = 6
Giaûi:
x2 + = x2 + = = (x + 0,25)2
	Vôùi x = 48,75 thì (49,75 + 0,25)2 = 502 = 2500
+ Khaùi quaùt hoùa baøi toaùn :
 1) Phaân tích ña thöùc x3m + 2 + x3n + 1 + 1 ( m ,n N ) thaønh nhaân töû 
 2) Cho ña thöùc : B = a4 + b4 + c4 – 2a2b2 – 2a2c2 – 2b2c2
	a) Phaân tích B thaønh boán nhaân töû baäc nhaát 
	b) Chöùng minh raèng neáu a , b , c laø soá ño ñoä daøi caùc caïnh cuûa moät tam giaùc thì b < 0
 3) Chöùng minh raèng vôùi moïi soá nguyeân n thì soá A = n3(n2 – 7)2 – 36n chia heát cho 105
+ Ñeà xuaát baøi taäp :
	1) Phaân tích caùc ña thöùc sau thaønh nhaân töû :
	a) x5 – x4 – x3 – x2 – x – 2 
	b) x8 + x6 + x4 + x2 + 1
	c) x8 + x7 + 1 
	d) x9 – x7 – x6 – x5 + x4 + x3 + x2 + x + 1 
	2) Phaân tích caùc ña thöùc sau thaønh nhaân töû baèng phöông phaùp ñaët aån phuï :
	a) (x2 + x)2 – 2(x2 + x) – 15 
	b) (x + 2)(x + 3)(x + 4)(x + 5) – 24 
	c) (x2 + 8x + 7)( x2 + 8x + 15) + 15
	d) (x2 + 3x + 1)( x2 + 3x + 2) – 6 
	3) Phaân tích caùc ña thöùc sau thaønh nhaân töû baèng phöông phaùp theâm , bôùt hoaëc taùch caùc haïng töû: 
	a) bc(b + c) + ca(c – a) – ab(a + b)
	b) 2a2b + 4ab2 – a2c + ac2 – 4b2c + 2bc2 – 4abc 
	c) y(x – 2z)2 + 8xyz + x(y – 2z)2 – 2z(x + y)2