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Đặt: $\sqrt{3x+y}=u,\sqrt{x+y}=v, u,v\ge0$.
Hệ trở thành:
$\left\{ \begin{array}{l} u+v=2\\v+u^2-2v^2=1 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} u=2-v\\ v+(2-v)^2-2v^2=1 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} u=2-v\\ v^2+3v-3=0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} v=\frac{-3+\sqrt{21}}{2}\\ u=\frac{7-\sqrt{21}}{2} \end{array} \right.$
Từ đó, suy ra: $(x,y)=(5-\sqrt{21},\frac{5-\sqrt{21}}{2})$
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