2-1.2n+4.2n=9.25
=>2n-1+22.2n=9.25
=>2n-1+2n+2=9.25
=>2n-1.(23+1)=9.25
=>2n-1.9=9.25
=>2n-1=25
=>n-1=5=>n=6
Ta có: \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Leftrightarrow2^n\cdot2^{-1}+2^n\cdot2^2=9\cdot2^5\)
\(\Leftrightarrow2^n\cdot\left(2^{-1}+2^2\right)=9\cdot2^5\)
\(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\)
\(\Leftrightarrow2^n=9\cdot2^5:\dfrac{9}{2}=2^5\cdot9\cdot\dfrac{2}{9}=2^6\)
hay n=6
Vậy: n=6