Đề sai thì phải ! Học Lớp 7 mới giải xong bài này !
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{9}\cdot\left(3^3\right)^n=3^n\)
\(\frac{1}{9}\cdot3^{3n}=3^n\)
\(\frac{1}{9}=3^n\text{ : }3^{3n}\)
\(\frac{1}{9}=3^{-2n}\)
\(\frac{1}{3^2}=\frac{1}{3^{2n}}\)
\(\Rightarrow\text{ }3^{2n}=3^2\)
\(3^{2n}-3^2=0\)
\(3\left(3^{2n-1}-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3=0\text{ ( Vô lí ) }\\3^{2n-1}-3=0\end{cases}}\) \(\Rightarrow\text{ }3^{2n-1}=3\) \(\Rightarrow\text{ }2n-1=1\) \(\Rightarrow\text{ }2n=2\) \(\Rightarrow\text{ }n=1\)
Vậy \(n=1\)
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\frac{3^{3n}}{3^2}=3^n\)
\(3^{3n}=3^2\cdot3^n\)
\(3^{3n}=3^{n+2}\)
\(\Rightarrow\text{ }3n=n+2\)
\(3n-n=2\)
\(2n=2\)
\(n=2\text{ : }2\)
\(n=1\)
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\frac{3^{3n}}{3^2}=3^n\)
\(3^{3n}=3^2\cdot3^n\)
\(3^{3n}=3^{n+2}\)
\(\Rightarrow\text{ }3n=n+2\)
\(3n-n=2\)
\(2n=2\)
\(n=2\text{ : }2\)
\(n=1\)
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\frac{3^{3n}}{3^2}=3^n\)
\(3^{3n}=3^2\cdot3^n\)
\(3^{3n}=3^{n+2}\)
\(\Rightarrow\text{ }3n=n+2\)
\(3n-n=2\)
\(2n=2\)
\(n=2\text{ : }2\)
\(n=1\)
