ĐKXĐ : \(x^2\ge y^2\)
P/t (2) <=> \(\dfrac{2y^2}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}=y\) \(\Leftrightarrow\left[{}\begin{matrix}y=0\\\dfrac{2y}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}=1\end{matrix}\right.\)
Với y = 0 thay vào p/t (1) : \(x^4=144\Leftrightarrow x=\sqrt[4]{144}\)
Với \(2y=\sqrt{x^2+y^2}+\sqrt{x^2-y^2}\) . Suy ra : \(\left\{{}\begin{matrix}\sqrt{x^2+y^2}=\dfrac{3y}{2}\\\sqrt{x^2-y^2}=\dfrac{y}{2}\end{matrix}\right.\)
Xét : \(\sqrt{x^2+y^2}=\dfrac{3y}{2}\Leftrightarrow x^2+y^2=\dfrac{9y^2}{4};y\ge0\) \(\Leftrightarrow x^2=\dfrac{5y^2}{4}\) ; \(y\ge0\)
Thay vào p/t (1) : \(\dfrac{9y^2}{4}.\dfrac{1}{4}y^2=144\Leftrightarrow y^4=256\Leftrightarrow y=\pm4\) ; y \(\ge0\Rightarrow y=4\)
\(\Rightarrow x=...\)
