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Đặt $a=x+y,b=\sqrt{x-y},b\ge 0$
Hệ trở thành: $\left\{ \begin{array}{l} a+b=8\\(a-b^2)b=4 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} b=8-a\\ (a-(8-a)^2)(8-a)=4 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} b=8-a\\ a^3-25a^2+200a-516=0 \end{array} \right.$
$\Leftrightarrow \left [ \begin{array}{l} \left\{ \begin{array}{l} a=6\\b=2 \end{array} \right.\\ \left\{ \begin{array}{l} a=\frac{1}{2}(19-\sqrt{17})\\ b=\frac{1}{2}(-3+\sqrt{17}) \end{array} \right. \end{array} \right.$
Từ đó suy ra: $(x;y)\in\{(5;1),(8-\sqrt{17};\frac{1}{2}(3+\sqrt{17}))\}$
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