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\(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\\ \Rightarrow2^n\left(\dfrac{1}{2}+4\right)=9\cdot2^5\\ \Rightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\\ \Rightarrow2^{n-1}\cdot9=9\cdot2^5\\ \Rightarrow n-1=5\\ \Rightarrow n=6\)
\(\Leftrightarrow2^n\left(\frac{1}{2}+4\right)=9\cdot2^5\Leftrightarrow2^n\cdot\frac{9}{2}=9\cdot2^5\Leftrightarrow2^n=2^6\Leftrightarrow n=6\)
\(2^{-1}.2^n+4.2^n=9.2^5\)
\(2^n.2=9.2^5\)
\(\Rightarrow2^n=9.2^4\)
Ko có n nhé bn
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
a) \(\frac{1}{9}.27^n=3^n\)
\(=>\frac{27^n}{9}=3^n\)
\(=>3^n=3^n=>n=1\)
b) \(2^{-1}.2^n+4.2^n=9.2^5\)
\(=>2^{n-1}.2^2.2^n=9.2^5\)
\(=>2^{n-1}.2^{2+n}=9.2^5\)
\(=>2^{2n+1}=9.5^2\)
\(=>n=\)
Câu b đề sai hay sao ấy số xấu lắm