Đáp án:
a. \(x \in \left( { - \infty ; - \dfrac{5}{2}} \right] \cup \left[ { - \dfrac{1}{2};\dfrac{1}{2}} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {2x + 5} \right)\left( {4{x^2} - 1} \right) \le 0\\
\to \left( {2x + 5} \right)\left( {2x - 1} \right)\left( {2x + 1} \right) \le 0
\end{array}\)
BXD:
x -∞ -5/2 -1/2 1/2 +∞
f(x) - 0 + 0 - 0 +
\(KL:x \in \left( { - \infty ; - \dfrac{5}{2}} \right] \cup \left[ { - \dfrac{1}{2};\dfrac{1}{2}} \right]\)
\(\begin{array}{l}
b.2{x^2} + 5x + 2 \le 3\sqrt {2{x^2} + 5x + 2} \\
\to \sqrt {2{x^2} + 5x + 2} \left( {\sqrt {2{x^2} + 5x + 2} - 3} \right) \le 0\\
\to \left\{ \begin{array}{l}
\sqrt {2{x^2} + 5x + 2} \ge 0\\
\sqrt {2{x^2} + 5x + 2} \le 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x + 2} \right)\left( {2x + 1} \right) \ge 0\\
2{x^2} + 5x + 2 \le 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ; - 2} \right] \cup \left[ { - \dfrac{1}{2}; + \infty } \right)\\
\left( {x - 1} \right)\left( {2x + 7} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ; - 2} \right] \cup \left[ { - \dfrac{1}{2}; + \infty } \right)\\
x \in \left[ { - \dfrac{7}{2};1} \right]
\end{array} \right.\\
KL:x \in \left[ { - \dfrac{7}{2}; - 2} \right] \cup \left[ { - \dfrac{1}{2};1} \right]
\end{array}\)
