Question

Giá trị của n thỏa mãn:\(\dfrac{8}{n^{2+1}}\)=\(\dfrac{32}{4^n}\)

Answer 1

ĐK: n ≠ 0
Ta co': \(\dfrac{8}{n^{2+1}}=\dfrac{32}{4^n}\) 
       \(\Rightarrow\dfrac{8}{n^3}=\dfrac{32:4}{4^n:4}\)
       \(\Rightarrow\dfrac{8}{n^3}=\dfrac{8}{4^{n-1}}\)
       \(\Rightarrow n^3=4^{n-1}\)
       \(\Rightarrow\left\{{}\begin{matrix}n=4\\3=n-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=4\\n=3+1=4\end{matrix}\right.\) \(\Rightarrow n=4\) (t/m)
Vậy \(n=4\)