Question

Bài 4

Cho \(2^x=8^{y+1}\)và \(9^y=3^{x-9}\) (x;y thuộc N ) Tính x+y

Answer 1

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 \(\left\{{}\begin{matrix}x=21\\y=6\end{matrix}\right.\)

Answer 2

\(2^x=8^{y+1}=\left(2^3\right)^{y+1}=2^{3y+3}\)

\(\Leftrightarrow x=3y+3\left(1\right)\)

\(9^y=3^{x-9}\)

\(\Leftrightarrow\left(3^2\right)^y=3^{x-9}\)

\(\Leftrightarrow3^{2y}=3^{x-9}\)

\(\Leftrightarrow2y=x-9\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow x+2y=3y+3+x-9\)

\(\Leftrightarrow x+y=2y+x-6\)