e)
\(\dfrac{x+2}{3}=\dfrac{3}{x+2}\\ =\left(x+2\right)^2=9\\ \left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\\ \left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
f)
\(\dfrac{x-4}{-5}=\dfrac{-5}{x-4}\\ \left(x-4\right)^2=25\\ \left[{}\begin{matrix}x-4=5\\x-4=-5\end{matrix}\right.\\ \left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
\(\dfrac{x+2}{3}=\dfrac{3}{x+2}\\ =>\left(x+2\right)^2=9\\ =>\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)