Đáp án:
$\begin{array}{l}
b)Dkxd:x \ge 0;x \ne 1\\
\left( {1 + \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{x\sqrt x + \sqrt x - x - 1}}} \right)\\
= \dfrac{{\sqrt x + 1 + \sqrt x }}{{\sqrt x + 1}}:\dfrac{{x + 1 - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{\left( {2\sqrt x + 1} \right)\left( {x + 1} \right)}}{{x - 1}}\\
c) - 1 < a < 1\\
\left( {\dfrac{1}{{\sqrt {1 + a} }} + \sqrt {1 - a} } \right):\left( {\dfrac{1}{{\sqrt {1 - {a^2}} }} + 1} \right)\\
= \dfrac{{1 + \sqrt {1 - a} .\sqrt {1 + a} }}{{\sqrt {1 + a} }}:\dfrac{{1 + \sqrt {1 - {a^2}} }}{{\sqrt {1 - {a^2}} }}\\
= \dfrac{{1 + \sqrt {1 - {a^2}} }}{{\sqrt {1 + a} }}.\dfrac{{\sqrt {1 + a} .\sqrt {1 - a} }}{{1 + \sqrt {1 - {a^2}} }}\\
= \sqrt {1 - a} \\
d)a \ge 0;a \ne 1\\
\dfrac{1}{{\sqrt a - 1}} - \dfrac{1}{{\sqrt a + 1}} + 1\\
= \dfrac{{\sqrt a + 1 - \left( {\sqrt a - 1} \right) + \left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{\sqrt a + 1 - \sqrt a + 1 + a - 1}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{a + 1}}{{a - 1}}\\
e)\left( {\dfrac{6}{{x - 9}} + \dfrac{1}{{\sqrt x + 3}}} \right).\dfrac{{x - 3\sqrt x }}{{\sqrt x }}\\
= \dfrac{{6 + \sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\sqrt x }}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x + 3}}\\
= 1
\end{array}$
