Đáp án: $\left( {x;y} \right) = \left( {0;\frac{5}{2}} \right)$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ne - 1;y \ne 2\\
\left\{ \begin{array}{l}
\frac{{x + 2}}{{x + 1}} + \frac{2}{{y - 2}} = 6\\
\frac{5}{{x + 1}} - \frac{1}{{y - 2}} = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{x + 1 + 1}}{{x + 1}} + 2.\frac{1}{{y - 2}} = 6\\
5.\frac{1}{{x + 1}} - \frac{1}{{y - 2}} = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
1 + \frac{1}{{x + 1}} + 2.\frac{1}{{y - 2}} = 6\\
5.\frac{1}{{x + 1}} - \frac{1}{{y - 2}} = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{1}{{x + 1}} + 2.\frac{1}{{y - 2}} = 5\\
5.\frac{1}{{x + 1}} - \frac{1}{{y - 2}} = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{1}{{x + 1}} + 2.\frac{1}{{y - 2}} = 5\\
10.\frac{1}{{x + 1}} - 2.\frac{1}{{y - 2}} = 6
\end{array} \right.\\
\Rightarrow 11.\frac{1}{{x + 1}} = 11\\
\Rightarrow \frac{1}{{x + 1}} = 1\\
\Rightarrow x + 1 = 1\\
\Rightarrow x = 0\left( {tmdk} \right)\\
Thay\,\frac{1}{{x + 1}} = 1\,vào\,5.\frac{1}{{x + 1}} - \frac{1}{{y - 2}} = 3\\
\Rightarrow \frac{1}{{y - 2}} = 2\\
\Rightarrow y - 2 = \frac{1}{2}\\
\Rightarrow y = \frac{5}{2}\\
Vậy\,\left( {x;y} \right) = \left( {0;\frac{5}{2}} \right)
\end{array}$
