Đáp án:
\(\left[ \begin{array}{l}
x = 1;y = 2\\
x = 2;y = 1
\end{array} \right.\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
2\left( {x + y} \right) - xy = 4\,\,\,\,\,\,\,\left( 1 \right)\\
xy\left( {x + y - 4} \right) = - 2\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow xy = 2\left( {x + y} \right) - 4\\
\left( 2 \right) \Leftrightarrow xy\left[ {\left( {x + y} \right) - 4} \right] = - 2\\
\Leftrightarrow \left[ {2\left( {x + y} \right) - 4} \right]\left[ {\left( {x + y} \right) - 4} \right] = - 2\\
\Leftrightarrow \left[ {\left( {x + y} \right) - 2} \right].\left[ {\left( {x + y} \right) - 4} \right] = - 1\\
\Leftrightarrow {\left( {x + y} \right)^2} - 6\left( {x + y} \right) + 8 = - 1\\
\Leftrightarrow {\left( {x + y} \right)^2} - 6\left( {x + y} \right) + 9 = 0\\
\Leftrightarrow {\left( {x + y - 3} \right)^2} = 0\\
\Leftrightarrow x + y = 3 \Rightarrow xy = 2\\
\left\{ \begin{array}{l}
x + y = 3\\
xy = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 3 - x\\
x\left( {3 - x} \right) = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 3 - x\\
{x^2} - 3x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1;y = 2\\
x = 2;y = 1
\end{array} \right.
\end{array}\)
