$\left\{\begin{matrix} \dfrac{4}{x + y} + \dfrac{1}{y - 1} = 5\\ \dfrac{1}{x + y} - \dfrac{2}{y - 1} = -1 \end{matrix}\right.$ (ĐK $x \neq 2; y \neq 1$)
Đặt $a = \dfrac{1}{x - 2}; b = \dfrac{1}{y - 1}$, hệ phương trình trở thành:
$\left\{\begin{matrix} a + b = 1\\ 2a - 3b = 1 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} 3a + 3b = 3\\ 2a - 3b = 1 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} 5a = 4\\ a + b = 1 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} a = \frac{4}{5}\\ \frac{4}{5} + b = 1 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} a = \frac{4}{5}\\ b = \frac{1}{5} \end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix} \dfrac{1}{x - 2} = \frac{4}{5}\\ \dfrac{1}{y - 1} = \frac{1}{5} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} x - 2 = \frac{5}{4}\\ y - 1 = 5 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} x = \frac{13}{4}\\ y = 6 \end{matrix}\right.$ (Thỏa mãn ĐK)
