Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 1\\
\sqrt {16x - 16} + \sqrt {9x - 9} - \sqrt {25x - 25} = 4\sqrt 2 \\
\Rightarrow \sqrt {16\left( {x - 1} \right)} + \sqrt {9\left( {x - 1} \right)} - \sqrt {25\left( {x - 1} \right)} = 4\sqrt 2 \\
\Rightarrow 4\sqrt {x - 1} + 3\sqrt {x - 1} - 5\sqrt {x - 1} = 4\sqrt 2 \\
\Rightarrow 2\sqrt {x - 1} = 4\sqrt 2 \\
\Rightarrow \sqrt {x - 1} = 2\sqrt 2 \\
\Rightarrow x - 1 = 8\\
\Rightarrow x = 9\left( {tmdk} \right)\\
Vay\,x = 9\\
b)Dkxd:x \ge 0;x \ne 9;x \ne 4\\
B = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 4\sqrt x + \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}
\end{array}$
