a) Ta có: \(\left(5-2x\right)^2+4x-10=8\)
⇔\(25-20x+4x^2+4x-10-8=0\)
\(\Leftrightarrow4x^2-16x+7=0\)
\(\Leftrightarrow4x^2-2x-14x+7=0\)
\(\Leftrightarrow2x\left(2x-1\right)-7\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{7}{2}\right\}\)
