\(3^{\left(n+2\right)\left(n-1\right)}=1\)
\(\Leftrightarrow3^{\left(n+2\right)\left(n-1\right)}=3^0\)
\(\Leftrightarrow\left(n+2\right)\left(n-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n+2=0\\n-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=1\end{matrix}\right.\)
Vậy ..
\(3^{\left(n+2\right)\left(n-1\right)}=1\Leftrightarrow\left(n+2\right)\left(n-1\right)=0\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=1\end{matrix}\right.\)
3(n + 2)(n - 1) = 1
⇔ 3(n + 2)(n - 1) = 30
⇔ (n + 2)(n - 1) = 0
⇔ \(\left[{}\begin{matrix}n+2=0\\n-1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}n=-2\\n=1\end{matrix}\right.\)