Giải thích các bước giải:
$\left\{ \begin{array}{l}\dfrac{2}{x+1}+\dfrac{6}{y-2}=5\\\dfrac{3}{x+1}-\dfrac{4}{y-2}=1\end{array} \right.\text{(I)}$
Thay $a=\dfrac{1}{x+1};b=\dfrac{1}{y-2}$ vào $\text{(I)}$, ta được:
$\left\{ \begin{array}{l}2a+6b=5\\3a-4b=1\end{array} \right.⇔\left\{ \begin{array}{l}6a+18b=15\\6a-8b=2\end{array} \right.\\⇔\left\{ \begin{array}{l}26b=13\\2a+6b=5\end{array} \right.⇔\left\{ \begin{array}{l}b=\dfrac{1}{2}\\2a+6.\dfrac{1}{2}=5\end{array} \right.\\⇔\left\{ \begin{array}{l}a=1\\b=\dfrac{1}{2}\end{array} \right.⇒\left\{ \begin{array}{l}\dfrac{1}{x+1}=1\\\dfrac{1}{y-2}=\dfrac{1}{2}\end{array} \right.\\⇒\left\{ \begin{array}{l}x+1=1\\y-2=2\end{array} \right.⇒\left\{ \begin{array}{l}x=0\\y=4\end{array} \right.$
Vậy $(x;y)=(0;4)$
