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đây là kq phân tích đa thức thành nhân tử
(x-1)*(2*x+1)*(x^2-x+2)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(x^4+x^3-4x^2+x+1\)
\(=\left(x^4-x^3\right)+\left(2x^3-2x^2\right)-\left(2x^2-2x\right)-\left(x-1\right)\)
\(=x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x^3-x^2\right)+\left(3x^2-3x\right)+\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)+3x\left(x-1\right)+\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x^2+3x+1\right)\)
\(=\left(x-1\right)^2\left(x^2+3x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+4x^3+2x^2-4x+1\)
\(=x^4+2x^3-x^2+2x^3+4x^2-2x-x^2-2x+1\)
\(=x^2\left(x^2+2x-1\right)+2x\left(x^2+2x-1\right)-\left(x^2+2x-1\right)\)
\(=\left(x^2+2x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)
\(=x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
b) Ta có: \(4x^4+y^4\)
\(=4x^4+y^4+4x^2y^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\)
a, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)
\(=x^2\left[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)\right]=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
b, \(4x^4+y^4=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)