Question

Giả hpt:\(\left\{{}\begin{matrix}\frac{1}{x+2}+\frac{2}{y-x}=3\\\frac{2}{x+2}-\frac{3}{y-x}=1\end{matrix}\right.\)

Answer 1

ĐK : \(x\ne-2\) , ta có :

\(\left\{{}\begin{matrix}\frac{1}{x+2}+\frac{2}{y-x}=3\\\frac{2}{x+2}-\frac{3}{y-x}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{x+2}+\frac{6}{y-x}=9\\\frac{4}{x+2}-\frac{6}{y-x}=2\end{matrix}\right.\)

\(\Rightarrow\frac{7}{x+2}=11\Leftrightarrow x+2=\frac{7}{11}\Leftrightarrow x=\frac{-15}{11}\)

Thay \(x=-\frac{15}{11}\)vào pt : \(\frac{1}{x+2}+\frac{2}{y-x}=3\) được :

\(\frac{1}{\frac{-15}{11}+2}+\frac{2}{y+\frac{15}{11}}=3\) \(\Leftrightarrow\frac{11}{7}+\frac{2}{y+\frac{15}{11}}=3\)

Giải ra được : \(y=\frac{2}{55}\)

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