Đáp án:
$\begin{array}{l}
g)\dfrac{{ - 3}}{{2 + x}} \ge 0\\
\Rightarrow 2 + x < 0\left( {do: - 3 < 0} \right)\\
\Rightarrow x < - 2\\
h){\left( {x + 5} \right)^2} \ge 0\left( {đúng\,mọi\,x} \right)\\
i)\dfrac{{\sqrt 6 - 4}}{{x + 2}} \ge 0\\
\Rightarrow x + 2 < 0\\
\Rightarrow x < - 2\\
j)x - 7 > 0\\
\Rightarrow x > 7\\
k)2x + 5 \ge 0\\
\Rightarrow x \ge - \dfrac{5}{2}\\
l)12x - 1 > 0\\
\Rightarrow x > \dfrac{1}{{12}}\\
m)\sqrt 5 - \sqrt 3 x \ge 0\\
\Rightarrow \sqrt 3 x \le \sqrt 5 \\
\Rightarrow x \le \dfrac{{\sqrt {15} }}{3}\\
n)\sqrt 6 x - 4x \ge 0\\
\Rightarrow \left( {\sqrt 6 - 4} \right).x \ge 0\\
\Rightarrow x \le 0\\
o)\left( {\sqrt x - 7} \right)\left( {\sqrt x + 7} \right) \ge 0\\
\Rightarrow x - 49 \ge 0\\
\Rightarrow x \ge 49\\
p){\left( {x - 6} \right)^6} \ge 0\left( {đúng\,mọi\,x} \right)\\
q) - 12x + 5 \ge 0\\
\Rightarrow x \le \dfrac{5}{{12}}\\
r)5x + 8 \ge 0\\
\Rightarrow x \ge \dfrac{{ - 8}}{5}\\
s)\dfrac{{ - 2\sqrt 6 + \sqrt {23} }}{{ - x + 5}} \ge 0\\
\Rightarrow - x + 5 < 0\\
\Rightarrow x > 5\\
t)2011 - {x^2} \ge 0\\
\Rightarrow {x^2} \le 2011\\
\Rightarrow - \sqrt {2011} \le x \le \sqrt {2011} \\
u)\dfrac{{2\sqrt {15} - \sqrt {59} }}{{x - 7}} \ge 0\\
\Rightarrow \dfrac{{\sqrt {60} - \sqrt {59} }}{{x - 7}} \ge 0\\
\Rightarrow x - 7 > 0\\
\Rightarrow x > 7\\
v)4{x^2} + 4x + 1 \ge 0\\
\Rightarrow {\left( {2x + 1} \right)^2} \ge 0\left( {mọi\,x} \right)\\
{\rm{w}})49{x^2} - 24x + 4 \ge 0\\
\Rightarrow {\left( {7x} \right)^2} - 2.7x.\dfrac{{12}}{7} + \dfrac{{144}}{{49}} + \dfrac{{52}}{{49}} \ge 0\\
\Rightarrow {\left( {7x - \dfrac{{12}}{7}} \right)^2} + \dfrac{{52}}{{49}} \ge 0\left( {luon\,đúng} \right)\\
y)\dfrac{{12x + 5}}{{\sqrt 3 }} \ge 0\\
\Rightarrow x \ge - \dfrac{5}{{12}}
\end{array}$
