Đáp án:
$\begin{array}{l}
B2)Dkxd:x \ge 1\\
2\sqrt {36x - 36} - \dfrac{1}{3}\sqrt {9x - 9} - 4\sqrt {4x - 4} + \sqrt {x - 1} = 16\\
\Leftrightarrow 2.6\sqrt {x - 1} - \dfrac{1}{3}.3\sqrt {x - 1} - 4.2\sqrt {x - 1} + \sqrt {x - 1} = 16\\
\Leftrightarrow 12\sqrt {x - 1} - \sqrt {x - 1} - 8\sqrt {x - 1} + \sqrt {x - 1} = 16\\
\Leftrightarrow 4\sqrt {x - 1} = 16\\
\Leftrightarrow \sqrt {x - 1} = 4\\
\Leftrightarrow x - 1 = 16\\
\Leftrightarrow x = 17\left( {tmdk} \right)\\
Vậy\,x = 17\\
B1)\\
5\sqrt 2 = \sqrt {25.2} = \sqrt {50} \\
2\sqrt 5 = \sqrt {{2^2}.5} = \sqrt {20} \\
2\sqrt 3 = \sqrt {{2^2}.3} = \sqrt {12} \\
3\sqrt 2 = \sqrt {{3^2}.2} = \sqrt {18} \\
\Leftrightarrow 2\sqrt 3 ;3\sqrt 2 ;2\sqrt 5 ;5\sqrt 2 \\
B2)\\
a)A = \left( {\sqrt 5 - 2} \right)\left( {\sqrt 5 + 2} \right)\\
= {\left( {\sqrt 5 } \right)^2} - {2^2}\\
= 5 - 4\\
= 1\\
b)B = \left( {\sqrt {45} + \sqrt {63} } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \sqrt 9 .\left( {\sqrt 5 + \sqrt 7 } \right)\left( {\sqrt 7 - \sqrt 5 } \right)\\
= 3.\left( {{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}} \right)\\
= 3.\left( {7 - 5} \right)\\
= 6
\end{array}$
