x^2-x-6 phân tích nhân tử
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)
\(=x\left(x-6\right)+2\left(x-6\right)=\left(x-6\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)\times\left(x+4\right)\times\left(x+6\right)\times\left(x+8\right)+16\)
\(=\left(x+2\right)\times\left(x+8\right)\times\left(x+4\right)\times\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\times\left(x^2+10x+24\right)+16\)
Đặt \(t=x^2+10x+16\), ta được :
\(t\times\left(t+8\right)+16\)
\(=t^2+8t+16\)
\(=\left(t+4^2\right)\)
Thay \(t=x^2+10x+16\), ta được :
\(\left(x^2+10x+16\right)^2\)
\(=\left[\left(x+2\right)\times\left(x+8\right)\right]^2\)
\(=\left(x+2\right)^2\times\left(x+8\right)^2\)
\(=\left(x+2\right)^2\left(x+8\right)^2\)
_ Vậy \(\left(x+2\right)\times\left(x+4\right)\times\left(x+6\right)\times\left(x+8\right)+16\)\(=\left(x+2\right)^2\left(x+8\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)-28\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28\)
\(=\left(x^2+5x\right)^2-64\)
\(=\left(x^2+5x+8\right)\left(x^2+5x-8\right)\)
![](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)
(3x+2)2-(x-6)2=(3x+2-x+6)(3x+2+x-6)=(2x+8)(4x-4)=8(x+4)(x-1)
\(^{x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)}\)
x^2-x-6=x^2-3x+2x-6
=(x^2-3x)+(2x-6)
=x(x-3)+2(x-3)
=(x+2).(x-3)