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Điều kiện: $x+y\ge0$.
Ta có: $\begin{cases}\sqrt{x^2+x+2}-\sqrt{x+y}=y\\\sqrt{x+y}=x-y+1 \end{cases} $
$\Leftrightarrow \left\{ \begin{array}{l}
\sqrt{x+y}+y=
\sqrt{x^2+x+2}\\
\sqrt{x+y}+y=x+1 \end{array} \right.$
Từ đó, suy ra: $\sqrt{x^2+x+2}=x+1$
$\Leftrightarrow \left\{ \begin{array}{l}
x^2+x+2=x^2+2x+1\\x\ge-1 \end{array} \right.$
$\Leftrightarrow x=1$
Dẫn tới: $\sqrt{y+1}=2-y$
$\Leftrightarrow \left\{ \begin{array}{l} y\le2\\ y+1=4-4y+y^2 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} y\le2\\y^2-5y+3=0 \end{array} \right.$
$y=\frac{1}{2}(5-\sqrt{13})$
Vậy: $(x,y)=(1,
\frac{1}{2}(5-\sqrt{13}) )$
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