Đáp án:
$\begin{array}{l}
b)\sqrt {3 - 4x} \\
Dkxd:3 - 4x \ge 0\\
\Rightarrow 4x \le 3\\
\Rightarrow x \le \dfrac{3}{4}\\
Vậy\,x \le \dfrac{3}{4}\\
c)Dkxd: - 2x + 6 \ge 0\\
\Rightarrow 2x \le 6\\
\Rightarrow x \le 3\\
Vậy\,x \le 3\\
d)\dfrac{2}{{x - 3}} \ge 0\\
\Rightarrow x - 3 > 0\\
\Rightarrow x > 3\\
Vậy\,x > 3\\
e)\dfrac{{2x + 6}}{3} \ge 0\\
\Rightarrow 2x + 6 \ge 0\\
\Rightarrow x \ge - 3\\
Vậy\,x \ge - 3\\
g)\dfrac{{2x + 1}}{4} \ge 0\\
\Rightarrow 2x + 1 \ge 0\\
\Rightarrow x \ge \dfrac{{ - 1}}{2}\\
Vậy\,x \ge - \dfrac{1}{2}\\
h)\dfrac{4}{{x + 2}} \ge 0\\
\Leftrightarrow x + 2 > 0\\
\Leftrightarrow x > - 2\\
Vậy\,x > - 2\\
B2)\\
1)5\sqrt {\dfrac{1}{5}} + \dfrac{1}{2}\sqrt {20} - \dfrac{5}{4}\sqrt {\dfrac{4}{5}} + \dfrac{1}{3}\sqrt {45} \\
= \dfrac{5}{{\sqrt 5 }} + \dfrac{1}{2}.2\sqrt 5 - \dfrac{5}{4}.\dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}.3\sqrt 5 \\
= \sqrt 5 + \sqrt 5 - \dfrac{1}{2}\sqrt 5 + \sqrt 5 \\
= \dfrac{{5\sqrt 5 }}{2}\\
2)\dfrac{{5 - \sqrt 5 }}{{1 - \sqrt 5 }} = \dfrac{{\sqrt 5 \left( {\sqrt 5 - 1} \right)}}{{1 - \sqrt 5 }} = - \sqrt 5 \\
3)\sqrt {{5^2}} + \sqrt {{{\left( { - 5} \right)}^2}.4} \\
= 5 + 5.2\\
= 15
\end{array}$
