Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2}y + x{y^2} = 0\left( 1 \right)\\
2{x^2} + 3xy + 2{y^2} = 1\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Rightarrow xy\left( {x + y} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
y = 0\\
x = - y
\end{array} \right.\\
+ Khi\,x = 0 \Rightarrow \left( 2 \right):2{y^2} = 1 \Rightarrow y = \pm \frac{{\sqrt 2 }}{2}\\
+ Khi:y = 0 \Rightarrow \left( 2 \right):2{x^2} = 1 \Rightarrow x = \pm \frac{{\sqrt 2 }}{2}\\
+ Khi:x = - y \Rightarrow \left( 2 \right):2{x^2} + 3x.\left( { - x} \right) + 2{x^2} = 1\\
\Rightarrow {x^2} = 1\\
\Rightarrow x = \pm 1 \Rightarrow y = \mp 1\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {0; \pm \frac{{\sqrt 2 }}{2}} \right);\left( { \pm \frac{{\sqrt 2 }}{2};0} \right);\left( { \pm 1; \mp 1} \right)} \right\}
\end{array}$
