Giải bpt $\frac{3(4x^{2}-9)}{\sqrt[]{3x^{2}-3} }$ $\leq$ 2x+3

Đáp án:

$\dfrac{3}{2} \le x \le 2$

Giải thích các bước giải:

$\begin{array}{l} DK:\left[ \begin{array}{l} x > 1\\ x <  - 1 \end{array} \right.\\ \dfrac{{3\left( {4{x^2} - 9} \right)}}{{\sqrt {3{x^2} - 3} }} \le 2x + 3\\  \to \dfrac{{3\left( {4{x^2} - 9} \right) - \left( {2x + 3} \right)\sqrt {3{x^2} - 3} }}{{\sqrt {3{x^2} - 3} }} \le 0\\  \to 3\left( {4{x^2} - 9} \right) - \left( {2x + 3} \right)\sqrt {3{x^2} - 3}  \le 0\left( {do:\sqrt {3{x^2} - 3}  > 0\forall \left[ \begin{array}{l} x > 1\\ x <  - 1 \end{array} \right.} \right)\\  \to 3\left( {2x - 3} \right)\left( {2x + 3} \right) - \left( {2x + 3} \right)\sqrt {3{x^2} - 3}  \le 0\\  \to \left( {2x + 3} \right)\left( {6x - 9 - \sqrt {3{x^2} - 3} } \right) \le 0\\ TH1:2x + 3 \le 0\\  \to x \le  - \dfrac{3}{2}\\  \to 6x - 9 - \sqrt {3{x^2} - 3}  \ge 0\\  \to 6x - 9 \ge \sqrt {3{x^2} - 3} \\  \to 36{x^2} - 108x + 81 \ge 3{x^2} - 3\left( {DK:x \ge \dfrac{3}{2}} \right)\\  \to \left[ \begin{array}{l} x \ge 2\\ x \le \dfrac{{14}}{{11}} \end{array} \right.\left( {KTM} \right)\\ TH2:2x + 3 \ge 0\\  \to x \ge  - \dfrac{3}{2}\\  \to 6x - 9 - \sqrt {3{x^2} - 3}  \le 0\\  \to 6x - 9 \le \sqrt {3{x^2} - 3} \\  \to 36{x^2} - 108x + 81 \le 3{x^2} - 3\left( {DK:x \ge \dfrac{3}{2}} \right)\\  \to \dfrac{{14}}{{11}} \le x \le 2\\ KL:\dfrac{3}{2} \le x \le 2 \end{array}$