ĐK: \(x\ne y\)
Biến đổi pt đầu:
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-6=0\Leftrightarrow x+y-\frac{6}{x-y}=0\)
\(\Leftrightarrow x+y-1-3.\frac{2}{x-y}=-1\)
Đặt \(\left\{{}\begin{matrix}x+y-1=a\\\frac{2}{x-y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-3b=-1\\a^2-b^2-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3b-1\\a^2-b^2-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left(3b-1\right)^2-b^2-3=0\)
\(\Leftrightarrow8b^2-6b-2=0\Rightarrow\left[{}\begin{matrix}b=1\Rightarrow a=2\\b=-\frac{1}{4}\Rightarrow a=-\frac{7}{4}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y-1=2\\\frac{2}{x-y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x-y=2\end{matrix}\right.\) \(\Rightarrow...\)
TH2: \(\left\{{}\begin{matrix}x+y-1=-\frac{7}{4}\\\frac{2}{x-y}=-\frac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-\frac{3}{4}\\x-y=-8\end{matrix}\right.\) \(\Rightarrow...\)
