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Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=a\\\frac{1}{y+2}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a+b=2\\8a-3b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{2}\\b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=\frac{1}{2}\\\frac{1}{y+2}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y+2=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
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