Question

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

Answer 1

(I) \(\left\{{}\begin{matrix}\frac{2}{x+y}+\frac{1}{x-y}\\\frac{1}{x+y}-\frac{3}{x-y}\end{matrix}\right.\)
đặt
\(\left\{{}\begin{matrix}\frac{1}{x+y}=a\\\frac{1}{x-y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a+b=3\\a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=3\\2a-6b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=3\\7b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{10}{7}\\b=\frac{1}{7}\end{matrix}\right.\)
với \(\left\{{}\begin{matrix}a=\frac{10}{7}\\b=\frac{1}{7}\end{matrix}\right.\)ta có :
\(\left\{{}\begin{matrix}\frac{1}{x+y}=\frac{10}{7}\\\frac{1}{x-y}=\frac{1}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10.\left(x+y\right)=7\\x-y=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{77}{20}\\y=\frac{-63}{20}\end{matrix}\right.\)
thử thôi ko bik đúng ko ạ