Đáp án:
$\begin{array}{l}
a)Đkxđ:x \ne 1;y \ne - 1\\
\left\{ \begin{array}{l}
\frac{x}{{x - 1}} + \frac{1}{{y + 1}} = 1\\
\frac{1}{{x - 1}} - \frac{y}{{y + 1}} = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{x - 1 + 1}}{{x - 1}} + \frac{1}{{y + 1}} = 1\\
\frac{1}{{x - 1}} - \frac{{y + 1 - 1}}{{y + 1}} = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
1 + \frac{1}{{x - 1}} + \frac{1}{{y + 1}} = 1\\
\frac{1}{{x - 1}} - 1 + \frac{1}{{y + 1}} = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{1}{{x - 1}} + \frac{1}{{y + 1}} = 0\\
\frac{1}{{x - 1}} + \frac{1}{{y + 1}} = 0
\end{array} \right.\\
Vậy\,hpt\,có\,nghiệm\,x \ne 1;y \ne - 1\,thỏa\,mãn\,\frac{1}{{x - 1}} + \frac{1}{{y + 1}} = 0
\end{array}$
Câu 2 tương tự câu 1 nhé
