Đáp án:
$\begin{array}{l}
B1)\\
a)Dkxd: - 2x + 3 \ge 0\\
\Leftrightarrow 2x \le 3\\
\Leftrightarrow x \le \dfrac{3}{2}\\
Vậy\,x \le \dfrac{3}{2}\\
b)\dfrac{4}{{x + 3}} \ge 0\\
\Leftrightarrow x + 3 > 0\\
\Leftrightarrow x > - 3\\
Vậy\,x > - 3\\
c)\dfrac{2}{{{{\left( {x - 1} \right)}^2}}} \ge 0\\
\Leftrightarrow x\# 1\\
Vậy\,x\# 1\\
d)\dfrac{{ - 1}}{{{x^2} + 1}} \ge 0\\
\Leftrightarrow {x^2} + 1 < 0\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
e)Dkxd:\dfrac{{2x + 1}}{3} \ge 0\\
\Leftrightarrow 2x + 1 \ge 0\\
\Leftrightarrow x \ge \dfrac{{ - 1}}{2}\\
Vậy\,x \ge \dfrac{{ - 1}}{2}\\
f)\dfrac{{ - 1}}{{2x + 3}} \ge 0\\
\Leftrightarrow 2x + 3 < 0\\
\Leftrightarrow x < \dfrac{{ - 3}}{2}\\
Vậy\,x < \dfrac{{ - 3}}{2}\\
B2)\\
a)\dfrac{{ - 5}}{{ - x - 7}} \ge 0\\
\Leftrightarrow - x - 7 < 0\\
\Leftrightarrow x > - 7\\
Vậy\,x > - 7\\
b)Dkxd:2x - 1 > 0\\
\Leftrightarrow x > \dfrac{1}{2}\\
Vậy\,x > \dfrac{1}{2}\\
c)Dk:\dfrac{{{x^2} + 1}}{{1 - 2x}} \ge 0\\
\Leftrightarrow 1 - 2x > 0\\
\Leftrightarrow x < \dfrac{1}{2}\\
Vậy\,x < \dfrac{1}{2}\\
d)2x - 1 > 0\\
\Leftrightarrow x > \dfrac{1}{2}\\
Vậy\,x > \dfrac{1}{2}\\
B3)\\
a)\sqrt {4 - 2\sqrt 3 } - \sqrt 3 \\
= \sqrt {3 - 2\sqrt 3 + 1} - \sqrt 3 \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt 3 \\
= \sqrt 3 - 1 - \sqrt 3 \\
= - 1\\
b)\sqrt {11 + 6\sqrt 2 } - 3 + \sqrt 2 \\
= \sqrt {9 + 2.3.\sqrt 2 + 2} - 3 + \sqrt 2 \\
= \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} - 3 + \sqrt 2 \\
= 3 + \sqrt 2 - 3 + \sqrt 2 \\
= 2\sqrt 2 \\
c)\sqrt {15 - 6\sqrt 6 } + \sqrt {33 - 12\sqrt 6 } \\
= \sqrt {9 - 2.3.\sqrt 6 + 6} - \sqrt {24 - 2.2\sqrt 6 .3 + 9} \\
= \sqrt {{{\left( {3 - \sqrt 6 } \right)}^2}} - \sqrt {{{\left( {2\sqrt 6 - 3} \right)}^2}} \\
= 3 - \sqrt 6 - 2\sqrt 6 + 3\\
= 6 - 3\sqrt 6
\end{array}$
