Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{x}{{x - 4}} + \dfrac{1}{{\sqrt x - 2}} + \dfrac{1}{{\sqrt x + 2}}\\
= \dfrac{x}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \dfrac{1}{{\sqrt x - 2}} + \dfrac{1}{{\sqrt x + 2}}\\
= \dfrac{{x + \left( {\sqrt x + 2} \right) + \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
b,\\
x = 25 \Rightarrow \sqrt x = 5 \Rightarrow A = \dfrac{5}{{5 - 2}} = \dfrac{5}{3}\\
c,\\
A = - \dfrac{1}{3} \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 2}} = - \dfrac{1}{3}\\
\Leftrightarrow 3\sqrt x = - 1.\left( {\sqrt x - 2} \right)\\
\Leftrightarrow 3\sqrt x = - \sqrt x + 2\\
\Leftrightarrow 4\sqrt x = 2\\
\Leftrightarrow \sqrt x = \dfrac{1}{5}\\
\Leftrightarrow x = \dfrac{1}{{25}}
\end{array}\)
