Ta có: \(2^{x+1}.\left(-3\right)^y=12^x\)
\(\Rightarrow2^{x+1}.\left(-3\right)^y=\left(3.4\right)^x\)
\(\Rightarrow2^{x+1}.\left(-3\right)^y=3^x.4^x\)
\(\Rightarrow2^{x+1}.\left(-3\right)^y=3^x.2^{2x}\)
\(\Rightarrow2^{x+1}.\left(-1\right)^y.3^y=3^x.2^{2x}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=y=1\end{matrix}\right.\)
Vậy x=1 , y=1
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