Đáp án:
\(y = - 1;x = \dfrac{{11}}{9}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 1;y \ne - 2\\
\left\{ \begin{array}{l}
\dfrac{{x + 1}}{{x - 1}} + \dfrac{{3y}}{{y + 2}} = 7\\
\dfrac{2}{{x - 1}} = 4 + \dfrac{5}{{y + 2}} = \dfrac{{4y + 8 + 5}}{{y + 2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{x + 1}}{{x - 1}} + \dfrac{{3y}}{{y + 2}} = 7\\
x - 1 = 2:\dfrac{{4y + 13}}{{y + 2}} = \dfrac{{2y + 4}}{{4y + 13}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2y + 4}}{{4y + 13}} + 1 = \dfrac{{2y + 4 + 4y + 13}}{{4y + 13}}\\
\dfrac{{x + 1}}{{x - 1}} + \dfrac{{3y}}{{y + 2}} = 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{6y + 17}}{{4y + 13}}\left( {y \ne - \dfrac{{13}}{4}} \right)\\
\dfrac{{\dfrac{{6y + 17}}{{4y + 13}} + 1}}{{\dfrac{{6y + 17}}{{4y + 13}} - 1}} + \dfrac{{3y}}{{y + 2}} = 7\left( * \right)
\end{array} \right.\\
\left( * \right) \to \dfrac{{6y + 17 + 4y + 13}}{{6y + 17 - 4y - 13}} + \dfrac{{3y}}{{y + 2}} = 7\\
\to \dfrac{{10y + 30}}{{2y + 4}} + \dfrac{{3y}}{{y + 2}} = 7\\
\to \dfrac{{10y + 30 + 6y - 7.2\left( {y + 2} \right)}}{{2\left( {y + 2} \right)}} = 0\\
\to 10y + 30 + 6y - 14y - 28 = 0\\
\to 2y + 2 = 0\\
\to y = - 1\left( {TM} \right)\\
\to x = \dfrac{{6\left( { - 1} \right) + 17}}{{4\left( { - 1} \right) + 13}} = \dfrac{{11}}{9}\left( {TM} \right)
\end{array}\)
