Đáp án:
\(0 \le x < 25\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{\sqrt x }}{{\sqrt x - 5}} + \dfrac{{10\sqrt x }}{{x - 25}} - \dfrac{5}{{\sqrt x + 5}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 5} \right) + 10\sqrt x - 5\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{x + 5\sqrt x + 10\sqrt x - 5\sqrt x + 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{x + 10\sqrt x + 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 5} \right)}^2}}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{\sqrt x + 5}}{{\sqrt x - 5}}\\
A < \dfrac{1}{3}\\
\to \dfrac{{\sqrt x + 5}}{{\sqrt x - 5}} < \dfrac{1}{3}\\
\to \dfrac{{3\sqrt x + 15 - \sqrt x + 5}}{{3\left( {\sqrt x - 5} \right)}} < 0\\
\to \dfrac{{2\sqrt x + 20}}{{3\left( {\sqrt x - 5} \right)}} < 0\\
\to \sqrt x - 5 < 0\left( {do:2\sqrt x + 20 > 0\forall x \ge 0} \right)\\
\to \sqrt x < 5\\
\to 0 \le x < 25
\end{array}\)
