Tìm số nguyên n:
\(\frac{1}{2}.2^n+4.2^n=9.2^5\)
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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
\(2^{-1}.2^n+4.2^n=9.2^5\)
\(2^n.2=9.2^5\)
\(\Rightarrow2^n=9.2^4\)
Ko có n nhé bn
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
\(\frac{1}{2}\cdot2^n+4.2^n=9.2^5\)
\(2^n\left(\frac{1}{2}+4\right)=9.32\)
\(2^n\left(\frac{1}{2}+\frac{8}{2}\right)=288\)
\(2^n\cdot\frac{9}{2}=288\)
\(2^n=288:\frac{9}{2}\)
\(2^n=288\cdot\frac{2}{9}\)
\(2^n=64\)
\(2^n=2^6\)
=> n = 6
vậy n = 6
Mk làm lun, ko viết lại đề bài nữa nhé =))
a) \(\Leftrightarrow\)\(3^2.3^{n+1}=9^4\)
\(\Leftrightarrow3^{n+1}=9^4:3^2\)
\(\Leftrightarrow3^{n+1}=3^6\)
\(\Rightarrow n+1=6\)
\(\Leftrightarrow n=6-1\)
\(\Rightarrow n=5\)
b)\(\Leftrightarrow2^n.\left(\frac{1}{2}+4\right)=9.2^5\)
\(\Leftrightarrow2^n.\frac{9}{2}=9.2^5\)
\(\Rightarrow2^n=\left(9.2^5\right):\frac{9}{2}\)
\(\Rightarrow2^n=468:\frac{9}{2}\)
Tự tính nốt KQ giúp mk nha ♥
a) \(\frac{1}{9}.27^n=3^n\)
\(\Leftrightarrow3^{-2}.3^{3n}=3^n\)
\(\Leftrightarrow3^{3n-2}=3^n\)
\(\Leftrightarrow3n-2=n\)
\(\Leftrightarrow2n=2\)
\(\Leftrightarrow n=1\)
b)\(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^{2+n}=3^7\)
\(\Leftrightarrow2+n=7\)
\(\Leftrightarrow n=5\)
\(\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(2^n.\left(\frac{1}{2}.4\right)=288\)
\(2^n.2=288\)
\(2^n=288:2\)
\(2^n=144\)
Suy ra n ko tìm được
Ta có :
\(\frac{1}{2}\cdot2^n+4\cdot2^n=\frac{9}{2}\cdot2^5\)
\(=>\left(\frac{1}{2}+4\right)\cdot2^n=9\cdot2^5\)
\(=>\left(\frac{1}{2}+\frac{8}{2}\right)\cdot2^n=9\cdot2^5\)
\(=>\frac{9}{2}\cdot2^n=\frac{9}{2}\cdot2^5\)
\(=>2^n=2^5\)
\(=>n=5\)