Question

\(\left\{{}\begin{matrix}2x+\frac{1}{x+y}=1\\8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

\(\left\{{}\begin{matrix}x+y+\frac{1}{x+y}+x-y=1\\5\left(x+y\right)^2+3\left(x-y\right)^2+\frac{5}{\left(x+y\right)^2}=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\frac{1}{x+y}+x-y=1\\5\left(x+y+\frac{1}{x+y}\right)^2+3\left(x-y\right)^2=23\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}a+b=1\\5a^2+3b^2=23\end{matrix}\right.\)

\(\Rightarrow5a^2+3\left(1-a\right)^2-23=0\)

\(\Leftrightarrow...\)