Đáp án:
\(\left[ \begin{array}{l}
y = 1\\
y = 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
x = 3 - y\\
\frac{{3 - y}}{y} + \frac{y}{{3 - y}} = \frac{5}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3 - y\\
\frac{{9 - 6y + {y^2} + {y^2}}}{{y\left( {3 - y} \right)}} = \frac{{5y\left( {3 - y} \right)}}{{2y\left( {3 - y} \right)}}\left( * \right)
\end{array} \right.\\
\left( * \right) \to \frac{{4{y^2} - 12y + 18}}{{2y\left( {3 - y} \right)}} = \frac{{ - 5{y^2} + 15y}}{{2y\left( {3 - y} \right)}}\left( {y \ne \left\{ {0;3} \right\}} \right)\\
\to 4{y^2} - 12y + 18 = - 5{y^2} + 15y\\
\to 9{y^2} - 27y + 18 = 0\\
\to {y^2} - 3y + 2 = 0\\
\to {y^2} - y - 2y + 2 = 0\\
\to \left( {y - 1} \right)\left( {y - 2} \right) = 0\\
\to \left[ \begin{array}{l}
y = 1\\
y = 2
\end{array} \right.\left( {TM} \right) \to \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\left( {TM} \right)
\end{array}\)
