Đặt \(\left\{{}\begin{matrix}\sqrt{x^2}=a\ge0\\\sqrt{y^2}=b\ge0\end{matrix}\right.\) hệ trở thành:
\(\left\{{}\begin{matrix}a=1-b^2\\b=1-a^2\end{matrix}\right.\) \(\Rightarrow a-b=a^2-b^2\)
\(\Leftrightarrow a-b=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}b=a\\b=1-a\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=1-a^2\\a=1-\left(1-a\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^2+a-1=0\\a^2-a=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=b=\frac{\sqrt{5}-1}{2}\\a=0\Rightarrow b=1\\a=1\Rightarrow b=0\end{matrix}\right.\)
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