Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
\dfrac{{a - \sqrt a }}{{1 - \sqrt a }} + \sqrt a = \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{1 - \sqrt a }} + \sqrt a = - \sqrt a + \sqrt a = 0\\
2,\\
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
y \ge 0\\
xy \ne 0\\
\sqrt x - \sqrt y \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
y > 0\\
x \ne y
\end{array} \right.\\
b,\\
A = \dfrac{{x\sqrt y + y\sqrt x }}{{\sqrt {xy} }}:\dfrac{1}{{\sqrt x - \sqrt y }}\\
= \dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt {xy} }}.\left( {\sqrt x - \sqrt y } \right)\\
= \left( {\sqrt x + \sqrt y } \right).\left( {\sqrt x - \sqrt y } \right)\\
= x - y\\
c,\\
x = \sqrt {3 + 2\sqrt 2 } = \sqrt {2 + 2.\sqrt 2 .1 + 1} = \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} = \sqrt 2 + 1\\
y = \sqrt {3 - 2\sqrt 2 } = \sqrt {2 - 2.\sqrt 2 .1 + 1} = \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = \sqrt 2 - 1\\
\Rightarrow A = x - y = \left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right) = 2
\end{array}\)
