Đáp án:
c. \(x \in \left[ { - \dfrac{1}{2};1} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\sqrt {{x^2} - 5x + 6} < 5 - x\\
\to \left\{ \begin{array}{l}
{x^2} - 5x + 6 \ge 0\\
5 - x \ge 0\\
{x^2} - 5x + 6 < 25 - 10x + {x^2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 3} \right) \ge 0\\
x \le 5\\
5x < 19
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ;2} \right] \cup \left[ {3; + \infty } \right)\\
x \le 5\\
x < \dfrac{{19}}{5}
\end{array} \right.\\
KL:x \in \left( { - \infty ;2} \right] \cup \left[ {3;\dfrac{{19}}{5}} \right)\\
2)\sqrt {{x^2} - 9} \ge x\\
\to \left\{ \begin{array}{l}
{x^2} - 9 \ge {x^2}\\
x \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 9 \ge 0\left( {vô lý} \right)\\
x \ge 0
\end{array} \right.
\end{array}\)
⇒ Bất phương tình vô nghiệm
\(\begin{array}{l}
c.\left| {x + 3} \right| \ge 2 + 2{x^2}\\
\to \left\{ \begin{array}{l}
2 + 2{x^2} \le x + 3\\
0 < 2 + 2{x^2}\left( {ld} \right)
\end{array} \right.\left( {do:2 + 2{x^2} > 0\forall x \in R} \right)\\
\to 2{x^2} - x - 1 \le 0\\
\to \left( {x - 1} \right)\left( {2x + 1} \right) \le 0\\
\to x \in \left[ { - \dfrac{1}{2};1} \right]
\end{array}\)
