1: \(\left|2x-1\right|=x^2-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)>0\\\left(x^2-4\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-4-2x+1\right)\left(x^2-4+2x-1\right)=0\\x\in\left(-\infty;-2\right)\cup\left(2;+\infty\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+1\right)\left(x^2+2x-5\right)=0\\x\in\left(-\infty;-2\right)\cup\left(2;+\infty\right)\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{3;-\sqrt{5}-1\right\}\)
2: Ta có: \(\left|x^2-5x+5\right|=2-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\\left(x^2-5x+5\right)^2=\left(2-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =2\\\left(x^2-5x+5-2+x\right)\left(x^2-5x+5+2-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =2\\\left(x^2-4x+3\right)\left(x^2-6x+7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{1;-\sqrt{2}+3\right\}\)
