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\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+2x^3+2x^2+2x+1\\ =\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\\ =x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+x^2+x+1\right)\left(x+1\right)\\ =\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left(x^2+1\right)\left(x+1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2y^2+2x^2+y^2+2\)
\(=x^2\left(y^2+2\right)+\left(y^2+2\right)\)
\(=\left(x^2+1\right)\left(y^2+2\right)\)
\(a^2-b^2+a-b\)
\(=\left(a+b\right)\left(a-b\right)+\left(a-b\right)\)
\(=\left(a+b+1\right)\left(a-b\right)\)
\(a,x^2y^2+2x^2+y^2+2\)
\(=y^2\left(x^2+1\right)+2\left(x^2+1\right)\)
\(=\left(y^2+2\right)\left(x^2+1\right)\)
\(b,a^2-b^2+a-b\)
\(=\left(a+b\right)\left(a-b\right)+\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(-x^2+2x+8\)
\(=-\left(x^2-2x-8\right)\)
\(=-\left(x-4\right)\left(x+2\right)\)
2: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2+3x-27\)
\(=2x^2+9x-6x-27\)
\(=x\left(2x+9\right)-3\left(2x+9\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-y^3+2x^2+2xy\)
\(=x\left(x^2-y^2+2x+2y\right)\)
\(=\)\(x\left[\left(x+y\right)\left(x-y\right)+2\left(x+y\right)\right]\)
\(=x\left(x+y\right)\left(x-y+2\right)\)