Lời giải:
Đặt $(\frac{1}{x+y-1}, \frac{1}{x-y+1})=(a,b)$ thì hpt trở thành:
\(\left\{\begin{matrix}
2a-4b=\frac{-14}{5}\\
3a+2b=\frac{-13}{5}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
a=-1\\
b=\frac{1}{5}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x+y-1=-1\\
x-y+1=5\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x+y=0\\ x-y=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=-2\end{matrix}\right.\)
\(ĐKXĐ:\left\{{}\begin{matrix}x+y\ne1\\x-y\ne-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x+y-1}\\b=\dfrac{1}{x-y+1}\end{matrix}\right.\)(1)
Khi đó HPT trở thành \(\left\{{}\begin{matrix}2a-4b=-\dfrac{14}{5}\\3a+2b=-\dfrac{13}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-2b=-\dfrac{7}{5}\\3a+2b=-\dfrac{13}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-2b=-\dfrac{7}{5}\\4a=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-2b=-\dfrac{7}{5}\\a=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{5}\\a=-1\end{matrix}\right.\)
Thay a = -1 ; b = 1/5 vào (1) ta được
\(\left\{{}\begin{matrix}-1=\dfrac{1}{x+y-1}\\\dfrac{1}{5}=\dfrac{1}{x-y+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\-2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)(TMĐK)
Vậy HPT có 1 nghiệm duy nhất (x;y) = (2;-2)
