c) \(\dfrac{\left(x+3\right)^5}{\left(x+3\right)^2}=\dfrac{64}{27}\)
\(\left(x+3\right)^3=\left(\dfrac{4}{3}\right)^3\)
\(x+3=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}-3\)
\(x=-\dfrac{5}{3}\)
\(c.\dfrac{\left(x+3\right)^5}{\left(x+3\right)^2}=\dfrac{64}{27}\\ \left(x+3\right)^5:\left(x+3\right)^2=\left(\dfrac{4}{3}\right)^3\\ \left(x+3\right)^3=\left(\dfrac{4}{3}\right)^3\\ x+3=\dfrac{4}{3}\\ x=\dfrac{4}{3}-3\\ x=\dfrac{4}{3}-\dfrac{9}{3}\\ x=-\dfrac{5}{3}\)
\(e.\left(\dfrac{4}{13}.\dfrac{6}{5}+\dfrac{4}{13}.\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\\ \left[\dfrac{4}{13}\left(\dfrac{6}{5}+\dfrac{2}{5}\right)\right]\left(2x+1\right)^2=\dfrac{10}{13}\\ \left(\dfrac{4}{13}.\dfrac{8}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\\ \dfrac{32}{65}.\left(2x+1\right)^2=\dfrac{10}{13}\\ \left(2x+1\right)^2=\dfrac{10}{13}:\dfrac{32}{65}\\ \left(2x+1\right)^2=\dfrac{10}{13}.\dfrac{65}{32}\\ \left(2x+1\right)^2=\dfrac{650}{416}\\ \left(2x+1\right)^2=\dfrac{25}{16}\\ \left(2x+1\right)^2=\left(\dfrac{5}{4}\right)^2\\ \left(2x+1\right)^2=\left(\pm\dfrac{5}{4}\right)^2\)
\(\left[{}\begin{matrix}2x+1=\dfrac{5}{4}\\2x+1=-\dfrac{5}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}2x=\dfrac{5}{4}-1\\2x=-\dfrac{5}{4}-1\end{matrix}\right.\\ \left[{}\begin{matrix}2x=\dfrac{5}{4}-\dfrac{4}{4}\\2x=-\dfrac{5}{4}-\dfrac{4}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}2x=\dfrac{1}{4}\\2x=-\dfrac{9}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{4}:2\\x=-\dfrac{9}{4}:2\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{4}.\dfrac{1}{2}\\x=-\dfrac{9}{4}.\dfrac{1}{2}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{8}\\x=-\dfrac{9}{8}\end{matrix}\right.\)