Đáp án:
$\begin{array}{l}
B1)\\
a)5 - 4\sqrt 3 + 3\sqrt {27} - \sqrt {7 - 4\sqrt 3 } \\
= 5 - 4\sqrt 3 + 3.3\sqrt 3 - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \\
= 5 + 5\sqrt 3 - \left( {2 - \sqrt 3 } \right)\\
= 5 + 5\sqrt 3 - 2 + \sqrt 3 \\
= 3 + 6\sqrt 3 \\
b)\dfrac{8}{{\sqrt {11} - \sqrt 3 }} + \dfrac{1}{{2 - \sqrt 3 }} - \dfrac{{11 - \sqrt {11} }}{{\sqrt {11} - 1}}\\
= \dfrac{{8\left( {\sqrt {11} + \sqrt 3 } \right)}}{{11 - 3}} + \dfrac{{2 + \sqrt 3 }}{{{2^2} - 3}} - \dfrac{{\sqrt {11} \left( {\sqrt {11} - 1} \right)}}{{\left( {\sqrt {11} - 1} \right)}}\\
= \sqrt {11} + \sqrt 3 + 2 + \sqrt 3 - \sqrt {11} \\
= 2\sqrt 3 + 2\\
B2)1)\sqrt {{x^2} - 6x + 9} = 3\\
\Leftrightarrow {x^2} - 6x + 9 = 9\\
\Leftrightarrow {x^2} - 6x = 0\\
\Leftrightarrow x\left( {x - 6} \right) = 0\\
\Leftrightarrow x = 0;x = 6\\
Vậy\,x = 0;x = 6\\
2)Dkxd:x \ge - 5\\
\sqrt {4x + 20} - 4\sqrt {x + 5} + \dfrac{3}{5}\sqrt {25x + 125} = 3\\
\Leftrightarrow 2\sqrt {x + 5} - 4\sqrt {x + 5} + \dfrac{3}{5}.5\sqrt {x + 5} = 3\\
\Leftrightarrow - 2\sqrt {x + 5} + 3\sqrt {x + 5} = 3\\
\Leftrightarrow \sqrt {x + 5} = 3\\
\Leftrightarrow x + 5 = 9\\
\Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4\\
B3)\\
a)Dkxd:x \ge 0;x \ne 1;x \ne 9\\
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{{2\sqrt x }}{{1 - x}}} \right):\dfrac{{\sqrt x - 3}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + 2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x - 1}}{{\sqrt x - 3}}\\
= \dfrac{{x - \sqrt x + 2\sqrt x }}{{\sqrt x + 1}}.\dfrac{1}{{\sqrt x - 3}}\\
= \dfrac{{x + \sqrt x }}{{\sqrt x + 1}}.\dfrac{1}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
b)x = 144\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 12\\
\Leftrightarrow P = \dfrac{{12}}{{12 - 3}} = \dfrac{4}{3}\\
c)P < 0\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 3}} < 0\\
\Leftrightarrow \sqrt x - 3 < 0\\
\Leftrightarrow \sqrt x < 3\\
\Leftrightarrow x < 9\\
Vậy\,0 \le x < 9;x \ne 1
\end{array}$
