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![](https://rs.olm.vn/images/avt/0.png?1311)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình k ghi lại đề nha bạn
\(=\left(x-y\right)^2-16z^2\\ =\left(x-y-4z\right)\left(x-y+4z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình nghĩ là đề thiếu đó bạn :)
đề đáng lẽ phải là: \(x^7+x^2+1\)
\(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left[x\left(x-1\right)\left(x+3\right)+1\right]\left(x^2+x+1\right)\)
\(=\left[\left(x^2-x\right)\left(x^3+1\right)+1\right]\left(x^2+x+1\right)\)
\(=\left(x^5-x^4-x^2-x+1\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài khó quá
4x⁴+4x-3= (2x)²+2(2x)+1-4
=(2x+1)²-2²=(2x+1-2)(2x+1+2)
=(2x-1)(2x+3)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)