Đáp án:
$\begin{array}{l}
1)a)\sqrt 9 + \sqrt {25} - \sqrt {36} \\
= 3 + 5 - 6\\
= 2\\
b)3\sqrt {48} - 2\sqrt {75} + 5\sqrt {27} \\
= 3.4\sqrt 3 - 2.5\sqrt 3 + 5.3\sqrt 3 \\
= 12\sqrt 3 - 10\sqrt 3 + 15\sqrt 3 \\
= 17\sqrt 3 \\
c)\dfrac{1}{{\sqrt 5 - 3}} - \dfrac{1}{{\sqrt 5 + 3}}\\
= \dfrac{{\sqrt 5 + 3 - \left( {\sqrt 5 - 3} \right)}}{{\left( {\sqrt 5 - 3} \right)\left( {\sqrt 5 + 3} \right)}}\\
= \dfrac{6}{{5 - 9}}\\
= \dfrac{{ - 3}}{2}\\
d)\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)\\
= {2^2} - 3\\
= 1\\
e)\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{\sqrt a - 1}}\left( {a \ge 0;a \ne 1} \right)\\
= \dfrac{{\sqrt a - 1 - \left( {\sqrt a + 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{ - 2}}{{a - 1}}\\
= \dfrac{2}{{1 - a}}\\
f)B = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{x - 2\sqrt x }}} \right).\left( {\sqrt x - 2} \right)\\
\left( {x > 0;x \ne 4} \right)\\
= \dfrac{{\sqrt x .\sqrt x + 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\left( {\sqrt x - 2} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x }}\\
= \sqrt x + 2\\
B2)\\
a)Dkxd:x \ge 0\\
\sqrt x = 5\\
\Rightarrow x = {5^2}\\
\Rightarrow x = 25\left( {tmdk} \right)\\
\text{Vậy}\,x = 25\\
b)Dkxd:x \ge 0\\
\sqrt {9x} - 5\sqrt x + 4\sqrt x = 6\\
\Rightarrow 3\sqrt x - 5\sqrt x + 4\sqrt x = 6\\
\Rightarrow 2\sqrt x = 6\\
\Rightarrow \sqrt x = 3\\
\Rightarrow x = 9\left( {tmdk} \right)\\
\text{Vậy}\,x = 9
\end{array}$
