Задача 21. Решите неравенство \(\dfrac{{4{x^2}-9}}{{\sqrt {3{x^2}-3} }} \leqslant \dfrac{2}{3}x + 1\)
ОТВЕТ: \(\left[ {-\dfrac{3}{2};\;-1} \right) \cup \left( {1;\;2} \right].\)
\(\dfrac{{4{x^2}-9}}{{\sqrt {3{x^2}-3} }} \le \dfrac{2}{3}x + 1\) Запишем область определения: \(3{x^2}-3 > 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left( {-\infty ;-1} \right) \cup \left( {1;\infty } \right).\) \(\dfrac{{4{x^2}-9}}{{\sqrt {3{x^2}-3} }} \le \dfrac{2}{3}x + 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\left( {2x-3} \right)\left( {2x + 3} \right)}}{{\sqrt {3{x^2}-3} }}-\dfrac{{2x + 3}}{3} \le 0\,\,\,\,\,\, \Leftrightarrow \) \(\dfrac{{3\left( {2x-3} \right)\left( {2x + 3} \right)-\left( {2x + 3} \right)\sqrt {3{x^2}-3} }}{{3\sqrt {3{x^2}-3} }} \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\left( {2x + 3} \right)\left( {6x-9-\sqrt {3{x^2}-3} } \right)}}{{3\sqrt {3{x^2}-3} }} \le 0.\) Решим данное неравенство методом интервалов. Найдём нули числителя: \(\left( {2x + 3} \right)\left( {6x-9-\sqrt {3{x^2}-3} } \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}2x + 3 = 0,\\\sqrt {3{x^2}-3} = 6x-9\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -\dfrac{3}{2},\\\left\{ \begin{array}{l}6x-9 \ge 0,\\3{x^2}-3 = {\left( {6x-9} \right)^2}\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -\dfrac{3}{2},\\\left\{ \begin{array}{l}x \ge \dfrac{3}{2},\\11{x^2}-36x + 28 = 0\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}x = -\dfrac{3}{2},\\\left\{ \begin{array}{l}x \ge \dfrac{3}{2},\\\left[ \begin{array}{l}x = 2,\\x = \dfrac{{14}}{{11}}\end{array} \right.\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -\dfrac{3}{2},\\x = 2.\end{array} \right.\) Найдём нули знаменателя: \(\sqrt {3{x^2}-3} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \pm 1.\) Ответ: \(\left[ {-\dfrac{3}{2};-1} \right) \cup \left( {1;2} \right].\) 