Đáp án:
$\begin{array}{l}
B1)\\
P = \left( {\dfrac{{2\sqrt 3 }}{{\sqrt 2 }} - 3\sqrt 3 + 5\sqrt 2 - \dfrac{1}{2}\sqrt 8 } \right).2\sqrt 6 - 5\sqrt 3 \\
= \left( {\dfrac{{2\sqrt 3 }}{{\sqrt 2 }} - 3\sqrt 3 + 5\sqrt 2 - 2\sqrt 2 } \right).2\sqrt 6 - 5\sqrt 3 \\
= \dfrac{{2\sqrt 3 }}{{\sqrt 2 }}.2\sqrt 6 - 3\sqrt 3 .2\sqrt 6 + 3\sqrt 2 .2\sqrt 6 - 5\sqrt 3 \\
= 12\sqrt 2 - 18\sqrt 2 + 12\sqrt 3 - 5\sqrt 3 \\
= 7\sqrt 3 - 6\sqrt 2 \\
B2)\\
a)Dkxd:x \ge 1\\
\sqrt {9\left( {x - 1} \right)} = 21\\
\Rightarrow 9\left( {x - 1} \right) = {21^2} = 441\\
\Rightarrow x - 1 = 49\\
\Rightarrow x = 50\left( {tmdk} \right)\\
b)Dkxd:x \ge 2\\
\sqrt {36x - 72} - 15\sqrt {\dfrac{{x - 2}}{{25}}} = 4\left( {5 + \sqrt {x - 2} } \right)\\
\Rightarrow 6\sqrt {x - 2} - 15.\dfrac{{\sqrt {x - 2} }}{5} = 20 + 4\sqrt {x - 2} \\
\Rightarrow 3\sqrt {x - 2} = 20 + 4\sqrt {x - 2} \\
\Rightarrow \sqrt {x - 2} = - 20\left( {ktm} \right)\\
\Rightarrow pt\,vo\,nghiem\\
c)\sqrt {{x^2} - 4x + 4} = 1 - x\left( {dkxd:x \le 1} \right)\\
\Rightarrow {x^2} - 4x + 4 = {x^2} - 2x + 1\\
\Rightarrow 2x = - 3\\
\Rightarrow x = - \dfrac{3}{2}\left( {tmdk} \right)
\end{array}$
