b) Ta có: \(\left\{{}\begin{matrix}\left|x^2-2x\right|\ge0\\\left|\left(x+1\right)\left(x-2\right)\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x^2-2x\right|+\left|\left(x+1\right)\left(x-2\right)\right|\ge0\)
Do đó, \(\left|x^2-2x\right|+\left|\left(x+1\right)\left(x-2\right)\right|=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left|x^2-2x\right|=0\\\left|\left(x+1\right)\left(x-2\right)\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x=0\\\left(x+1\right)\left(x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)=0\\\left(x+1\right)\left(x-2\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\\\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)