Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
k,\\
D = R\\
{x^2} + 2\sqrt {{x^2} - 3x + 11} \le 3x + 4\\
\Leftrightarrow \left( {{x^2} - 3x - 4} \right) + 2\sqrt {{x^2} - 3x + 11} \le 0\\
\Leftrightarrow \left( {{x^2} - 3x + 11} \right) + 2\sqrt {{x^2} - 3x + 11} - 15 \le 0\\
\Leftrightarrow \left( {\sqrt {{x^2} - 3x + 11} + 5} \right)\left( {\sqrt {{x^2} - 3x + 11} - 3} \right) \le 0\\
\sqrt {{x^2} - 3x + 11} + 5 > 0,\,\,\,\forall x\\
\Rightarrow \sqrt {{x^2} - 3x + 11} - 3 \le 0\\
\Leftrightarrow {x^2} - 3x + 11 \le 9\\
\Leftrightarrow {x^2} - 3x + 2 \le 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right) \le 0 \Leftrightarrow 1 \le x \le 2\\
\Rightarrow S = \left[ {1;2} \right]\\
i,\\
D = \left( { - \infty ;\dfrac{{24}}{5}} \right]\\
x \ge \sqrt {24 - 5x} \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} \ge 24 - 5x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} + 5x - 24 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left( {x + 8} \right)\left( {x - 3} \right) \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
x \ge 3\\
x \le - 8
\end{array} \right.
\end{array} \right. \Leftrightarrow x \ge 3\\
\Rightarrow S = \left[ {3;\dfrac{{24}}{5}} \right]\\
m,\\
\left| {{x^2} - 1} \right| = {x^2} - 2x + 8\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 2x + 8 \ge 0\\
\left[ \begin{array}{l}
{x^2} - 1 = {x^2} - 2x + 8\\
{x^2} - 1 = - {x^2} + 2x - 8
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\forall x\\
\left[ \begin{array}{l}
2x = 9\\
2{x^2} - 2x + 7 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \dfrac{9}{2}
\end{array}\)
