\(\left(x-5\right)^2=\left(1-3x\right)^2\\ =>\left[{}\begin{matrix}x-5=1-3x\\x-5=-\left(1-3x\right)\end{matrix}\right.\\ =>\left[{}\begin{matrix}x+3x=1+5\\x-5=-1+3x\end{matrix}\right.\\ =>\left[{}\begin{matrix}4x=6\\-2x=4\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
\(\left(x-5\right)^2=\left(1-3x\right)^2\)
TH1:
\(x-5=1-3x\)
\(\Rightarrow x+3x=1+5\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{3}{2}\)
TH2:
\(x-5=-\left(1-3x\right)\)
\(\Rightarrow x-5=-1+3x\)
\(\Rightarrow x-3x=-1+5\)
\(\Rightarrow-2x=4\)
\(\Rightarrow x=-2\)
