a, \(\frac{8}{2^n}=2\Rightarrow2.2^n=8\)
\(\Rightarrow2^{n+1}=2^3\)
\(\Rightarrow n+1=3\)
\(\Rightarrow n=2\)
d,\(\left(2n-3\right)^2=9\)
\(\left(2n-3\right)^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}2n-3=-3\\2n-3=3\end{cases}\Rightarrow\orbr{\begin{cases}2n=-3+3\\2n=3+3\end{cases}\Rightarrow}\orbr{\begin{cases}2n=0\\2n=6\end{cases}\Rightarrow}\orbr{\begin{cases}n=0\\n=3\end{cases}}}\)
Vậy n=0; n= 3