a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\x\ne y\end{matrix}\right.\)
b) Ta có: \(A=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\frac{\sqrt{x}-\sqrt{y}}{1}\)
\(=x-y\)
c) Ta có: \(x=\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left|\sqrt{2}+1\right|\)
\(=\sqrt{2}+1\)(Vì \(\sqrt{2}>1>0\))(nhận)
Ta có: \(y=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left|\sqrt{2}-1\right|\)
\(=\sqrt{2}-1\)(Vì \(\sqrt{2}>1\))(nhận)
Thay \(x=\sqrt{2}+1\) và \(y=\sqrt{2}-1\) vào biểu thức A=x-y, ta được:
\(A=\sqrt{2}+1-\left(\sqrt{2}-1\right)\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
\(=2\)
Vậy: Khi \(x=\sqrt{3+2\sqrt{2}}\) và \(y=\sqrt{3-2\sqrt{2}}\) thì A=2
