a) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)

b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)

a) (x³-x²)+(8x-8)=x(x-1)+8(x-1)=(x²+8)(x-1)

b) 8x³-8x²y+2xy²=2x(4x²-4xy+y²)

c) (x²+y²-z²)² - 4x²y²=(x²+y²-z²)² - (2xy)²=(x²+y²-z²-2xy)(x²+y²-z²+2xy)

a: \(=5a\left(x-2y\right)\)

b: \(=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

c: =(x-1)(x-7)

a)\(5ax-10ay=5a\left(x-2y\right)\)

b) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x+1\right)\left(x-y\right)\)

c) \(x^2-8x+7=\left(x-7\right)\left(x-1\right)\)

a)5a(x-2y)

b)\(\left(x^2+x\right)-\left(xy+y\right)\)

\(x\left(x+1\right)-y\left(x=1\right)\)

\(\left(x+1\right)\left(x-y\right)\)

c)\(x^2-x-7x+7\)

\(x\left(x-1\right)-7\left(x-1\right)\)

\(\left(x-1\right)\left(x-7\right)\)

a: \(x^4-4x^3-8x^2+8x\)

\(=x\left(x^3-4x^2-8x+8\right)\)

\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)

\(=x\left(x+2\right)\left(x^2-6x+4\right)\)

b: \(x^2-1-xy+y\)

\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)

\(=\left(x-1\right)\left(x-y+1\right)\)

c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)

\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)

\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)

Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!

Trả lời:

A=(x-1)(x+2)(x-3)(x+4)-144

A= (x²-5x-14)(x²-5x-24)-144 (1)

đặt m=x²-5x-14

=> A= m.(m-10)-144

A=m²-10m-144

A= (m-18)(m+8)

thay m vào, ta có:

A= (x²-5x-32)(x²-5x-6)

A=(x²-5x-32)(x+1)(x-6)

c: \(=\left(5x-y\right)\left(5x+y\right)\)

e: \(=\left(x-2\right)\left(x-3\right)\)

a) x(4y-10x)

b)3(x+2y)+(x+1)

c)(5x-y)(5x+y)

d)5x(y-z)²

e)(x-3)(x-2)

f)(2x+y)³

a) 

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

\(a,=2x^2-6x+9x-27=\left(x-3\right)\left(2x+9\right)\\ b,=x^2-7x+\dfrac{49}{4}-\dfrac{73}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{73}{4}=\left(x-\dfrac{7}{2}-\dfrac{\sqrt{73}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{\sqrt{73}}{2}\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ d,=x^2-2x-8x+16=\left(x-2\right)\left(x-8\right)\\ e,=x^2-3x-5x+15=\left(x-3\right)\left(x-5\right)\\ g,=x^2+2x+4x+8=\left(x+2\right)\left(x+4\right)\)

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(t=x^2+5x+4\)

(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)

Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)

a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+15\right)\left(3x-4\right)\)

a: 6x-2y=2(3x-y)

b: =(x-y)(x-2)(x+2)

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

b: =(x-y)(x-2)(x+2)