Đáp án:
x=2; y=1
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0;y \ne - 1\\
\left\{ \begin{array}{l}
2x + 3y = xy + 5\\
\dfrac{1}{x} + \dfrac{1}{{y + 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x + 3y = xy + 5\\
\dfrac{1}{x} = 1 - \dfrac{1}{{y + 1}} = \dfrac{{y + 1 - 1}}{{y + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x + 3y = xy + 5\\
\dfrac{1}{x} = \dfrac{y}{{y + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x + 3y = xy + 5\\
x = \dfrac{{y + 1}}{y}\left( {y \ne 0} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{2y + 2}}{y} + 3y = \dfrac{{{y^2} + y}}{y} + 5\left( * \right)\\
x = \dfrac{{y + 1}}{y}
\end{array} \right.\\
\left( * \right) \to \dfrac{{2y + 2 + 3{y^2}}}{y} = \dfrac{{{y^2} + y + 5y}}{y}\\
\to 2y + 2 + 3{y^2} = {y^2} + y + 5y\\
\to 2{y^2} - 4y + 2 = 0\\
\to {y^2} - 2y + 1 = 0\\
\to {\left( {y - 1} \right)^2} = 0\\
\to y = 1\left( {TM} \right)\\
\to x = 2\left( {TM} \right)
\end{array}\)
