ĐKXĐ x≥0,y≥0
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{5x}-\sqrt{2y}=\sqrt{10}\left(1\right)\\\sqrt{2x}+y=2\left(2\right)\end{matrix}\right.\) Từ (2) ⇒ \(y=2-\sqrt{2x}\)
Thay vào (1) ta được:
\(\Rightarrow\sqrt{5x}-\sqrt{2\left(2-\sqrt{2x}\right)}=\sqrt{10}\)\(\Leftrightarrow\sqrt{5x}-\sqrt{4-2\sqrt{2x}}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{4-2\sqrt{2x}}=\sqrt{5x}-\sqrt{10}\)
\(\Rightarrow4-2\sqrt{2x}=5x+10-10\sqrt{2x}\) (Bình phương 2 vế )
\(\Leftrightarrow-2\sqrt{2x}+10\sqrt{2x}=5x+10-4\)
\(\Leftrightarrow8\sqrt{2x}=5x+6\) \(\Rightarrow\left(8\sqrt{2x}\right)^2=\left(5x+6\right)^2\) \(\Leftrightarrow128x=25x^2+60x+36\)
\(\Leftrightarrow25x^2-68x+36=0\) \(\Leftrightarrow\left(25x^2-50x\right)-\left(18x-36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(25x-18\right)=0\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{18}{25}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=2-\sqrt{2\cdot2}=0\left(TM\right)\\y=2-\sqrt{2\cdot\dfrac{18}{25}}=\dfrac{4}{5}\left(TM\right)\end{matrix}\right.\)
