Giải thích các bước giải:

 Bài 3:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 7x + 10 \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
\left( {x - 2} \right)\left( {2{x^2} + 5x + 3} \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow {x^2} - 7x + 10 \le 0 \Leftrightarrow \left( {x - 2} \right)\left( {x - 5} \right) \le 0 \Leftrightarrow 2 \le x \le 5\\
\left( 2 \right) \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 > 0\\
2{x^2} + 5x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 < 0\\
2{x^2} + 5x + 3 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
\left( {2x + 3} \right)\left( {x + 1} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2\\
\left( {2x + 3} \right)\left( {x + 1} \right) < 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
\left[ \begin{array}{l}
x >  - 1\\
x <  - \frac{3}{2}
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2\\
 - \frac{3}{2} < x <  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > 2\\
 - \frac{3}{2} < x <  - 1
\end{array} \right.\\
 \Rightarrow 2 < x \le 5\\
 \Rightarrow S = \left( {2;5} \right]
\end{array}\)

Bài 4:

 TXĐ:  \(D = \left[ { - 5;3} \right]\)

Ta có:

\(\begin{array}{*{20}{l}}
{\sqrt {\left( {x + 5} \right)\left( {3 - x} \right)} {\rm{ }} \le {x^2} + 2x + m,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall x \in \left[ { - 5;3} \right]}\\
{{\rm{n}} \Leftrightarrow \sqrt {3x - {x^2} + 15 - 5x} {\rm{ }} - {x^2} - 2x \le m,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall x \in \left[ { - 5;3} \right]}\\
{{\rm{n}} \Leftrightarrow \sqrt { - {x^2} - 2x + 15} {\rm{ }} + \left( { - {x^2} - 2x + 15} \right) \le m + 15,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall x \in \left[ { - 5;3} \right]{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( 1 \right)}\\
{t = \sqrt { - {x^2} - 2x + 15} {\rm{ }} \Rightarrow t \ge 0,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall x \in \left[ { - 5;3} \right]}\\
{{t^2} = {\rm{ }} - {x^2} - 2x + 15 = {\rm{ }} - {{\left( {x + 1} \right)}^2} + 16 \le 16,{\mkern 1mu} {\mkern 1mu} \forall x}\\
{{\rm{n}} \Rightarrow t \le 4,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall x \in \left[ { - 5;3} \right]}\\
{{\rm{n}} \Rightarrow t \in \left[ {0;4} \right]}\\
{\left( 1 \right) \Leftrightarrow {t^2} + t \le m + 15,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall t \in \left[ {0;4} \right]}\\
{{\rm{n}} \Leftrightarrow m \ge {t^2} + t - 15,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \forall t \in \left[ {0;4} \right]}\\
{{\rm{n}} \Leftrightarrow m \ge \mathop {\max }\limits_{\left[ {0;4} \right]} f\left( t \right) = {t^2} + t - 15 = {4^2} + 4 - 15 = 5}\\
{{\rm{n}} \Rightarrow m \ge 5}
\end{array}\)

Vậy \(m \in \left[ {5; + \infty } \right)\)