Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\,\,\,\,\,{x^2} - 5x + 6 \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne 3
\end{array} \right.\\
\frac{{2{x^2} + 3x - 2}}{{{x^2} - 5x + 6}} \ge 0\\
\Leftrightarrow \frac{{\left( {2x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \le - 2\\
\frac{1}{2} \le x < 2\\
x > 3
\end{array} \right.\\
d,\,\,\,\,\,\,{x^2} - 3x - 10 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge 5\\
x \le - 2
\end{array} \right.\\
\sqrt {{x^2} - 3x - 10} < x - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3x - 10 < {\left( {x - 2} \right)^2}\\
x - 2 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3x - 10 < {x^2} - 4x + 4\\
x > 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x < 14\\
x > 2
\end{array} \right. \Rightarrow 5 \le x < 14
\end{array}\)
