Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
{x^2} + \left| {3x - 10} \right| - 10 > 0\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,3x - 10 \ge 0 \Leftrightarrow x \ge \frac{{10}}{3}\\
\left( 1 \right) \Leftrightarrow {x^2} + 3x - 10 - 10 > 0\\
\Leftrightarrow {x^2} + 3x - 20 > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > \frac{{ - 3 + \sqrt {89} }}{2}\\
x < \frac{{ - 3 - \sqrt {89} }}{2}
\end{array} \right. \Rightarrow x \ge \frac{{10}}{3}\\
\Rightarrow {S_1} = \left[ {\frac{{10}}{3}; + \infty } \right)\\
TH2:\,\,3x - 10 < 0 \Leftrightarrow x < \frac{{10}}{3}\\
\left( 1 \right) \Leftrightarrow {x^2} + \left( {10 - 3x} \right) - 10 > 0\\
\Leftrightarrow {x^2} - 3x > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 3\\
x < 0
\end{array} \right.\\
\Rightarrow {S_2} = \left( { - \infty ;0} \right) \cup \left( {3;\frac{{10}}{3}} \right)\\
S = {S_1} \cup {S_2} = \left( { - \infty ;0} \right) \cup \left( {3; + \infty } \right)\\
d,\\
{\rm{DK:}}\,\,\,\,\,{\rm{x}} \ge \frac{1}{2}\\
\left( {{x^2} - x - 6} \right)\sqrt {2x - 1} \ge 0\\
\Leftrightarrow {x^2} - x - 6 \ge 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x = \frac{1}{2}
\end{array} \right.
\end{array}\)
