Question

Tìm m,n\(\in\)Z:

\(2^{-1}\times2^n+4\times2^n=9\times2^5\)

\(2^m-2^n=1984\)

\(\dfrac{1}{9}\times27^n=3^n\)

\(\left(\dfrac{4}{9}\right)^n=\left(\dfrac{3}{2}\right)^{-5}\)

\(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)

\(-\dfrac{512}{343}=\left(-\dfrac{8}{7}\right)^n\)

Answer 1

a) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot2^5\)

\(\Rightarrow2^n\cdot4,5=288\)

\(\Rightarrow2^n=64\)

\(\Rightarrow n=6\)

b) \(2^m-2^n=1984\)

\(\Rightarrow2^n\cdot\left(2^{m-n}-1\right)=2^6\cdot31\)

\(\Rightarrow\left\{{}\begin{matrix}2^n=2^6\\2^{m-n}-1=31\end{matrix}\right.\)

\(\Rightarrow n=6\)

\(\Rightarrow2^{m-n}=32\Rightarrow m-n=5\Rightarrow m=11\)