câu 1:
a) Ta có: \(\left(3x-5\right)^2=\left(x+1\right)^2\)
\(\Rightarrow\left(3x-5\right)^2-\left(x+1\right)^2=0\)
\(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\4x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\4x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;3\right\}\)