Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 25;x \ne 9\\
B = \left( {\dfrac{{x - 5\sqrt x }}{{x - 25}} - 1} \right):\left( {\dfrac{{25 - x}}{{x + 2\sqrt x - 15}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x - 5}}{{\sqrt x - 3}}} \right)\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} - 1} \right)\\
:\dfrac{{25 - x - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x + 5}} - 1} \right).\dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}{{25 - x - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + x - 25}}\\
= \dfrac{{\sqrt x - \sqrt x - 5}}{{\sqrt x + 5}}.\dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}{{ - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{1}.\dfrac{1}{{ - \left( {\sqrt x + 3} \right)}}\\
= \dfrac{5}{{\sqrt x + 3}}\\
b)B < 1\\
\Leftrightarrow \dfrac{5}{{\sqrt x + 3}} - 1 < 0\\
\Leftrightarrow \dfrac{{5 - \sqrt x - 3}}{{\sqrt x + 3}} < 0\\
\Leftrightarrow 2 - \sqrt x < 0\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > 4\\
Vậy\,x > 4;x \ne 9;x \ne 25
\end{array}$
