Ta có: \(\left|x\left(x-7\right)\right|=-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-7\right)=-x\left(1\right)\\-x\left(x-7\right)=-x\left(2\right)\end{matrix}\right.\)
* Phương trình (1):
\(x\left(x-7\right)=-x\)
\(\Leftrightarrow x^2-7x=-x\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
* Phương trình (2):
\(-x\left(x-7\right)=-x\)
\(\Leftrightarrow x\left(x-7\right)=x\)
\(\Leftrightarrow x^2-7x=x\)
\(\Leftrightarrow x^2-8x=0\)
\(\Leftrightarrow x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
Vậy: \(x=0;6;8\)
Vì \(\left|x\left(x-7\right)\right|\ge0\\ \Rightarrow-x\ge0\) hay \(x\le0\)
Vậy x = 0
