a/ \(\left(2x-5\right)^2+2\left(2x-5\right)-8=0\)
Đặt \(2x-5=t\)
\(\Rightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=2\\2x-5=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{7}{2}\end{matrix}\right.\)
b/ Đặt \(x^2+2x+1=\left(x+1\right)^2=t\ge0\)
\(\Rightarrow\left(t+2\right)t=3\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+2x+1=1\Leftrightarrow x\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
