Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
A = \frac{x}{{x - 4}} + \frac{1}{{\sqrt x - 2}} + \frac{1}{{\sqrt x + 2}}\\
= \frac{x}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{x + \sqrt x + 2 + \sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{x + 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{\sqrt x }}{{\sqrt x - 2}}
\end{array}\)
b,
Thay x=25 ta được:
\[A = \frac{{\sqrt {25} }}{{\sqrt {25} - 2}} = \frac{5}{{5 - 2}} = \frac{5}{3}\]
c,
\[\begin{array}{l}
A = \frac{{ - 1}}{3} \Leftrightarrow \frac{{\sqrt x }}{{\sqrt x - 2}} = - \frac{1}{3}\\
\Leftrightarrow - 3\sqrt x = \sqrt x - 2\\
\Leftrightarrow \sqrt x = \frac{1}{2}\\
\Rightarrow x = \frac{1}{4}
\end{array}\]
