Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{\sqrt x }}{{x - 4}} + \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x + \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x + 2}}{{x - 4}}\\
b,\\
x = 4\,\,\,\,\left( {L,\,\,\,x \ge 0,x \ne 4} \right)\\
x = \sqrt {6 + 2\sqrt 5 } - \sqrt {6 - 2\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \left( {\sqrt 5 + 1} \right) - \left( {\sqrt 5 - 1} \right)\\
= 2\\
\Rightarrow A = \dfrac{{2\sqrt 2 + 2}}{{2 - 4}} = \dfrac{{2\sqrt 2 + 2}}{{ - 2}} = - \left( {\sqrt 2 + 1} \right)\\
2,\\
B = \dfrac{{2\sqrt x }}{{\sqrt x - 3}} + \dfrac{{x + 9\sqrt x }}{{x - 9}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x - 3}} + \dfrac{{x + 9\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) + \left( {x + 9\sqrt x } \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3x + 15\sqrt x }}{{x - 9}}
\end{array}\)
