Do \(\dfrac{4^x}{2^{x+y}}=8\)
\(\Rightarrow4^x=8.2^{x+y}\)
\(\Rightarrow\left(2^2\right)^x=2^3.2^{x+y}\)
\(\Rightarrow2^{2x}=2^{x+y+3}\)
\(\Rightarrow2x=x+y+3\)
\(\Rightarrow x=y+3^{\left(1\right)}\)
Mà \(\dfrac{9^{x+y}}{3^{5y}}=243\)
\(\Rightarrow9^{x+y}=243.3^{5y}\)
\(\Rightarrow\left(3^2\right)^{x+y}=3^5.3^{5y}\)
\(\Rightarrow3^{2x+2y}=3^{5+5y}\)
\(\Rightarrow2x+2y=5+5y\)
\(\Rightarrow2x=5+3y^{\left(2\right)}\)
Từ (1) và (2) suy ra: 2(y+3)=5+3y
\(\Rightarrow2y+6=5+3y\)
\(\Rightarrow3y-2y=6-5\)
\(\Rightarrow y=1\)
Thay y=1 vào (1) ta được: x=1+3=4
Vậy x=3;y=1
