Đáp án:
$\begin{array}{l}
1)a)\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \sqrt 3 - 1 + \sqrt 3 + 1\\
= 2\sqrt 3 \\
b)\sqrt {50} + \sqrt {72} - \sqrt {32} \\
= 5\sqrt 2 + 6\sqrt 2 - 4\sqrt 2 \\
= 7\sqrt 2 \\
c)\dfrac{2}{{\sqrt 5 + 2}} + \dfrac{2}{{\sqrt 5 - 2}}\\
= \dfrac{{2\left( {\sqrt 5 - 2} \right) + 2\left( {\sqrt 5 + 2} \right)}}{{5 - 4}}\\
= 4\sqrt 5 \\
d)\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} = \dfrac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }} = - \sqrt 7 \\
e)\sqrt {25a} + \sqrt {49a} - \sqrt {64a} \\
= 5\sqrt a + 7\sqrt a - 8\sqrt a \\
= 4\sqrt a \\
B2)a)Dkxd:x \ge 0\\
5\sqrt {12x} - 4\sqrt {3x} + 2\sqrt {48x} = 14\\
\Leftrightarrow 5.2\sqrt {3x} - 4\sqrt {3x} + 2.4\sqrt {3x} = 14\\
\Leftrightarrow 14\sqrt {3x} = 14\\
\Leftrightarrow \sqrt {3x} = 1\\
\Leftrightarrow 3x = 1\\
\Leftrightarrow x = \dfrac{1}{3}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{3}\\
b)\sqrt {{{\left( {x - 1} \right)}^2}} = \dfrac{1}{4}\\
\Leftrightarrow {\left( {x - 1} \right)^2} = \dfrac{1}{{16}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = \dfrac{1}{4}\\
x - 1 = - \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{4}\\
x = \dfrac{3}{4}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{4};x = \dfrac{3}{4}\\
B3)A = \sqrt {2 - \sqrt 3 } - \sqrt {2 + \sqrt 3 } \\
\Leftrightarrow {A^2} = 2 - \sqrt 3 - 2\sqrt {2 - \sqrt 3 } .\sqrt {2 + \sqrt 3 } + 2 + \sqrt 3 \\
= 4 - 2\sqrt {4 - 3} \\
= 4 - 2\\
= 2\\
\Leftrightarrow A = - \sqrt 2 \left( {do:A < 0} \right)
\end{array}$
