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bạn tham khảo 1 số bài rồi tự làm nhé
a) 3x−3+5(1−x)
=3x−3+5−5x
=3x−5x+2
=x(3−5)+2
=−2x+2
=2(1−x)
b) 12a2−3ab+8ac−2bc
=3a(4a−b)+2c(4a−b)
=(4a−b)(3a+2c)
c) x2−25+y2−2xy
=x2−2xy+y2−25
=(x−y)2−52
=(x−y−5)(x−y+5)
![](https://rs.olm.vn/images/avt/0.png?1311)
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
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a) \(4\left(x^2-y^2\right)+4x+1\)
\(=4x^2-4y^2+4x+1\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1^2\right]-\left(2y\right)^2\)
\(=\left(2x+1\right)^2-\left(2y\right)^2\)
\(=\left(2x-2y+1\right)\left(2x+2y+1\right)\)
b) \(x^2+1-x^3-x^2\)
\(=1-x^3\)
\(=\left(1-x\right)\left(1+x+x^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4y^4+4x^2y^2+4-4x^2y^2=\left(x^2y^2+2\right)^2-\left(2xy\right)^2=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)
\(4x^4y^4+4x^2y^2+1-4x^2y^2=\left(2x^2y^2+1\right)^2-\left(2xy\right)^2=\left(2x^2y^2-2xy+1\right)\left(2x^2y^2+2xy+1\right)\)
1. \(x^4y^4\)+4= \(\left(x^2y^2\right)^2\)+4\(x^2y^2\)+4-4\(x^2y^2\)=(\(x^2y^2\)+2)^2 - 4x^2y^2=(x^2y^2+2-2xy)(X^2y^2+2+2xy) .Câu b tương tự bạn tự làm nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x^4+4=x^4+4x^2+4-2x^4=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
\(b,4x^8+1=4x^8+4x^4+1-4x^4\)
\(=\left(2x^4+1\right)^2-4x^4=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)
\(c,4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-4x^2y^2\)
\(=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(=64y^4+32xy^3+8y^2x^2-32xy^3-16x^2y^2-4x^3y+8x^2y^2+4x^3y+x^4\)
\(=8y^2\left(8y^2+4xy+x^2\right)-4xy\left(8y^2+4xy+x^2\right)+x^2\left(8y^2+4xy+x^2\right)\)
\(=\left(8y^2-4xy+x^2\right)\left(8y^2+4xy+x^2\right)\)
b/ \(=y^4+2xy^3+2x^2y^2-2xy^3-4x^2y^2-4x^3y+2x^2y^2+4x^3y+4x^4\)
\(=y^2\left(y^2+2xy+2x^2\right)-2xy\left(y^2+2xy+2x^2\right)+2x^2\left(y^2+2xy+2x^2\right)\)
\(=\left(y^2-2xy+2x^2\right)\left(y^2+2xy+2x^2\right)\)
c/ \(=x^4+5x^3+7x^2+5x^3+25x^2+35x+3x^2+15x+21\)
\(=x^2\left(x^2+5x+7\right)+5x\left(x^2+5x+7\right)+3\left(x^2+5x+7\right)\)
\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)
d/ \(=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2-1-2xy\)
\(=\left(x-y\right)^2-1\)
\(=\left(x-y+1\right)\left(x-y-1\right)\)
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a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
a) 4x4 + 1
= (2x2)2 + 4x2 + 1 - 4x2 (Thêm và bớt đi 4x2 tổng ko thay đổi và để tạo ra hằng đẳng thức)
= (2x2 + 1)2 - (2x)2
= (2x2 + 1 - 2x)(2x2 + 1 + 2x)
b) 4x4 + y4
= (2x2)2 + 4x2y2 + (y2)2 - 4x2y2 (Cũng thêm và bớt đi tương tự như a)
= (2x2 + y2)2 - (2xy)2
= (2x2 + y2 - 2xy)(2x2 + y2 + 2xy)
c) x4 + 324
= (x2)2 + 36x2 + 182 - 36x2 (324 = 182)
= (x2 + 18)2 - (6x)2
= (x2 + 18 - 6x)(x2 + 18 + 6x)
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