Ta có : \(2^x=8^{y+1}\Rightarrow2^x=2^{3y+3}\)
\(\Rightarrow x=3y+3\)
Và \(9^y=3^{x-y}\Rightarrow3^{2y}=3^{x-9}\)
\(\Rightarrow2y=x-9\)
Thay \(x=3y+3\) , thì ta lại có:
\(2y=3y+3-9\)
\(\Rightarrow2y=3y-6\)
\(\Rightarrow2y-3y=-6\)
\(\Rightarrow-y=-6\Rightarrow y=6\)
Thay \(y=6\) vào thì ta có : \(x=3y+3\)
\(\Rightarrow x=3.6+3=21\)
Vậy \(\left\{{}\begin{matrix}x=21\\y=6\end{matrix}\right.\)
\(2^x=8^{y+1}=\left(2^3\right)^{y+1}=2^{3y+3}\)
\(\Leftrightarrow x=3y+3\left(1\right)\)
\(9^y=3^{x-9}\)
\(\Leftrightarrow\left(3^2\right)^y=3^{x-9}\)
\(\Leftrightarrow3^{2y}=3^{x-9}\)
\(\Leftrightarrow2y=x-9\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow x+2y=3y+3+x-9\)
\(\Leftrightarrow x+y=2y+x-6\)
