ĐKXĐ: \(x;y\ge\frac{1}{2}\)
Trừ vế cho vế ta được:
\(\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}-\sqrt{2-\frac{1}{y}}=0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}+\frac{x-y}{xy\left(\sqrt{2-\frac{1}{x}}+\sqrt{2-\frac{1}{y}}\right)}=0\)
\(\Leftrightarrow\frac{x-y}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x-y}{xy\left(\sqrt{2-\frac{1}{x}}+\sqrt{2-\frac{1}{y}}\right)}=0\)
\(\Leftrightarrow\left(x-y\right)\left(\frac{1}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{1}{xy\left(\sqrt{2-\frac{1}{x}}+\sqrt{2-\frac{1}{y}}\right)}\right)=0\)
\(\Leftrightarrow x=y\) (phần ngoặc to phía sau luôn dương)
Thay vào pt đầu:
\(\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}=2\) \(\Rightarrow\) đặt \(\frac{1}{\sqrt{x}}=a\) pt trở thành:
\(\sqrt{2-a^2}=2-a\Leftrightarrow\left\{{}\begin{matrix}a\le2\\2-a^2=\left(2-a\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\le2\\2a^2-4a+2=0\end{matrix}\right.\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
