Đáp án:
\(x \in \left[ {\dfrac{{ - 35 - \sqrt {685} }}{{18}}; - \dfrac{3}{2}} \right) \cup \left( { - 1;\dfrac{{ - 35 + \sqrt {685} }}{{18}}} \right] \cup \left[ {0;1} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { - \dfrac{3}{2}; - 1;1;\dfrac{3}{2}} \right\}\\
\dfrac{{6x\left( {2{x^2} + 5x + 3} \right) - x\left( {2{x^2} - 5x + 3} \right)}}{{\left( {2x + 3} \right)\left( {x + 1} \right)\left( {x - 1} \right)\left( {2x - 3} \right)}} \ge 0\\
\to \dfrac{{12{x^3} + 30{x^2} + 18x - 3{x^3} + 5{x^2} - 3x}}{{\left( {2x + 3} \right)\left( {x + 1} \right)\left( {x - 1} \right)\left( {2x - 3} \right)}} \ge 0\\
\to \dfrac{{9{x^3} + 35{x^2} + 15x}}{{\left( {2x + 3} \right)\left( {x + 1} \right)\left( {x - 1} \right)\left( {2x - 3} \right)}} \ge 0\\
\to \dfrac{{x\left( {9{x^2} + 35x + 15} \right)}}{{\left( {2x + 3} \right)\left( {x + 1} \right)\left( {x - 1} \right)\left( {2x - 3} \right)}} \ge 0\\
Xét:9{x^2} + 35x + 15 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 35 + \sqrt {685} }}{{18}}(N1)\\
x = \dfrac{{ - 35 - \sqrt {685} }}{{18}}(N2)
\end{array} \right.
\end{array}\)
BXD:
x -∞ N2 -3/2 -1 N1 0 1 3/2 +∞
f(x) - 0 + // - // + 0 - 0 + // - // +
\(KL:x \in \left[ {\dfrac{{ - 35 - \sqrt {685} }}{{18}}; - \dfrac{3}{2}} \right) \cup \left( { - 1;\dfrac{{ - 35 + \sqrt {685} }}{{18}}} \right] \cup \left[ {0;1} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)\)
