Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
b \ge 0\\
\sqrt a - \sqrt b \ne 0\\
\sqrt {ab} \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
b > 0\\
a \ne b
\end{array} \right.\\
b,\\
A = \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2} - 4\sqrt {ab} }}{{\sqrt a - \sqrt b }} - \dfrac{{a\sqrt b + b\sqrt a }}{{\sqrt {ab} }}\\
= \dfrac{{\left( {a + 2\sqrt {ab} + b} \right) - 4\sqrt {ab} }}{{\sqrt a - \sqrt b }} - \dfrac{{{{\sqrt a }^2}.\sqrt b + {{\sqrt b }^2}.\sqrt a }}{{\sqrt {ab} }}\\
= \dfrac{{a - 2\sqrt {ab} + b}}{{\sqrt a - \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a + \sqrt b } \right)}}{{\sqrt {ab} }}\\
= \dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{\sqrt a - \sqrt b }} - \left( {\sqrt a + \sqrt b } \right)\\
= \left( {\sqrt a - \sqrt b } \right) - \left( {\sqrt a + \sqrt b } \right)\\
= - 2\sqrt b ,\,\,\,\forall a\\
2,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x\sqrt x - 1 \ne 0\\
\sqrt x - 1 \ne 0\\
x + \sqrt x + 1 \ne 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
b,\\
B = \left( {\dfrac{{2\sqrt x + x}}{{x\sqrt x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 1}}{{x + \sqrt x + 1}}\\
= \left( {\dfrac{{2\sqrt x + x}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{\left( {2\sqrt x + x} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{1}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{1}{{x - 1}}
\end{array}\)
