Question

\(\left\{{}\begin{matrix}\dfrac{2}{x-y}+\dfrac{6}{y+x}=1,1\\\dfrac{4}{x-y}-\dfrac{9}{y+x}=0,1\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(x\ne\pm y\)

\(\left\{{}\begin{matrix}\dfrac{2}{x-y}+\dfrac{6}{x+y}=1,1\\\dfrac{4}{x-y}-\dfrac{9}{x+y}=0,1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{4}{x-y}+\dfrac{12}{x+y}=2,2\\\dfrac{4}{x-y}-\dfrac{9}{x+y}=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{x+y}=2,1\\\dfrac{4}{x-y}-\dfrac{9}{x+y}=0,1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y=10\\\dfrac{4}{x-y}=0,1+\dfrac{9}{10}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=14\\x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)