Giải thích các bước giải:
ĐKXĐ: $\left( {x + y} \right) \ne \left\{ {0; - 5; - 32} \right\}$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = \dfrac{{\left( {x - 5} \right)\left( {y + 10} \right)}}{{x + y + 5}}\\
\dfrac{{xy}}{{x + y}} = \dfrac{{\left( {x + 40} \right)\left( {y - 8} \right)}}{{x + y + 32}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = \dfrac{{xy + 10x - 5y - 50}}{{x + y + 5}}\\
\dfrac{{xy}}{{x + y}} = \dfrac{{xy - 8x + 40y - 320}}{{x + y + 32}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = \dfrac{{xy + 10x - 5y - 50}}{{x + y + 5}} = \dfrac{{xy - \left( {xy + 10x - 5y - 50} \right)}}{{x + y - \left( {x + y + 5} \right)}}\\
\dfrac{{xy}}{{x + y}} = \dfrac{{xy - 8x + 40y - 320}}{{x + y + 32}} = \dfrac{{xy - \left( {xy - 8x + 40y - 320} \right)}}{{x + y - \left( {x + y + 32} \right)}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = \dfrac{{ - 10x + 5y + 50}}{{ - 5}}\\
\dfrac{{xy}}{{x + y}} = \dfrac{{8x - 40y + 320}}{{ - 32}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = 2x - y - 10\\
\dfrac{{xy}}{{x + y}} = \dfrac{{ - 1}}{4}x + \dfrac{5}{4}y - 10
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = 2x - y - 10\\
2x - y - 10 = \dfrac{{ - 1}}{4}x + \dfrac{5}{4}y - 10
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{xy}}{{x + y}} = 2x - y - 10\\
x = y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{{x^2}}}{{2x}} = 2x - x - 10\\
x = y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{2} = x - 10\\
x = y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 20\\
x = y
\end{array} \right.\\
\Leftrightarrow x = y = 20
\end{array}$
Vậy hệ có nghiệm duy nhất là: $(x;y)=(20;20)$
