![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=4x\left(x+2\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=x^2\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2-9y^2+4x-6y=\left(4x^2-9y^2\right)+\left(4x-6y\right)=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(4x^2-9y^2+4x-6y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(1 + x2)2 - 4x(1 - x2)
= (1 + x2)(1 + x2) - 4x(1 - x2)
= (1 + x2 - 4x)(1 + x2 - 1 + x2)
= 2x2(x2 - 4x + 1)
Ta có: \(\left(x^2+1\right)^2+4x\left(x^2-1\right)\)
\(=x^4+2x^2+1+4x^3-4x\)
\(=x^4+2x^3+2x^3+4x^2-2x^2-4x+1\)
\(=\left(x+2\right)\left(x^3+2x^2-2x\right)+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x^2+5x+8\right)\left(x^2+4x+2x+8\right)=\left(x^2+5x+8\right)\left[x\left(x+4\right)+2\left(x+4\right)\right]\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)
\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)
\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
\(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
4x^2 + 8x
= 4x . x + 4x . 2
= 4x ( x + 2 )
\(4x^{^{ }2}+8x\)
\(=4x.x+4x.2\)
\(=4x\left(x+2\right)\)