Ta có :
\(\left(3n+2\right)^4=\left(3n+2\right)^6\)
\(\Leftrightarrow\left(3n+2\right)^6-\left(3n+2\right)^4=0\)
\(\Leftrightarrow\left[\left(3n+2\right)^2-1\right]-\left(3n+4\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(3n+2\right)^2-1\right]=0\\\left(3n+2\right)^4=0\end{matrix}\right.\)
+)\(\left(3n+2\right)^4=0\)
\(\Leftrightarrow n=\dfrac{2}{3}\)\(\left(tm\right)\)
+) \(\left[\left(3n+2\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3n+2=1\\3n+2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)\(\left(tm\right)\)
Vậy ....................
Ta có: \(\left(3n+2\right)^4=\left(3n+2\right)^6\)
\(\Leftrightarrow\left[{}\begin{matrix}3n+2=1\\3n+2=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3n=1-2\\3n=-1-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3n=-1\\3n=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)