\(-\frac{32}{\left(-2\right)^n}=4\\ =>-\frac{32}{\left(-2\right)^n}=\frac{4}{1}\\ =>\left(-2\right)^n=8\\ =>n=3\)
\(-\frac{32}{\left(-2\right)^n}=4\Rightarrow\frac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)5:(-2)n=(-2)2
=> 5-n=2
=> n=3
\(\frac{-32}{\left(-2\right)^n}=4\Leftrightarrow\frac{-32}{-8}=4\Rightarrow\left(-2\right)^n=4\Rightarrow n=2\)
\(-\frac{32}{\left(-2\right)^n}=4\)
\(\Rightarrow\frac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^5:\left(-2\right)^n=\left(-2\right)^2\)
\(\Rightarrow5-n=2\)
\(\Rightarrow n=5-2\)
\(\Rightarrow n=3\)
Vậy : \(n=3\)