Đáp án:
\[\left( {x;y} \right) = \left\{ {\left( { - \frac{4}{5};\frac{2}{5}} \right);\left( {0;0} \right);\left( {1;1} \right)} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + xy - 2{y^2} = 0\,\,\,\,\,\,\,\,\left( 1 \right)\\
3x + 2y = 5xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow {x^2} + xy - 2{y^2} = 0\\
\Leftrightarrow \left( {{x^2} - xy} \right) + \left( {2xy - 2{y^2}} \right) = 0\\
\Leftrightarrow x\left( {x - y} \right) + 2y\left( {x - y} \right) = 0\\
\Leftrightarrow \left( {x - y} \right)\left( {x + 2y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - y = 0\\
x + 2y = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = y\\
x = - 2y
\end{array} \right.\\
TH1:\,\,\,\,\,x = y\\
\left( 2 \right) \Leftrightarrow 3x + 2x = 5x.x\\
\Leftrightarrow 5{x^2} - 5x = 0\\
\Leftrightarrow {x^2} - x = 0\\
\Leftrightarrow x\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = y = 0\\
x = y = 1
\end{array} \right.\\
TH2:\,\,\,\,x = - 2y\\
3.\left( { - 2y} \right) + 2y = 5.\left( { - 2y} \right).y\\
\Leftrightarrow - 6y + 2y = - 10{y^2}\\
\Leftrightarrow 10{y^2} - 4y = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = 0 \Rightarrow x = 0\\
y = \frac{2}{5} \Rightarrow x = - \frac{4}{5}
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( { - \frac{4}{5};\frac{2}{5}} \right);\left( {0;0} \right);\left( {1;1} \right)} \right\}
\end{array}\)
