\(\dfrac{{2x + 3}}{{\sqrt {6{x^2} + 7x-3} }} \ge 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{2x + 3-2\sqrt {6{x^2} + 7x-3} }}{{\sqrt {6{x^2} + 7x-3} }} \ge 0.\)
Решим данное неравенство методом интервалов при условии, что:
\(6{x^2} + 7x-3 > 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left( {-\infty ;-\dfrac{3}{2}} \right) \cup \left( {\dfrac{1}{3};\infty } \right).\)
Найдём нули числителя:
\(2x + 3-2\sqrt {6{x^2} + 7x-3} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\sqrt {6{x^2} + 7x-3} = 2x + 3\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{4\left( {6{x^2} + 7x-3} \right) = {{\left( {2x + 3} \right)}^2},}\\{2x + 3 \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = \dfrac{7}{{10}},\\x = -\dfrac{3}{2},\end{array} \right.\\x \ge -\dfrac{3}{2}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{7}{{10}},\\x = -\dfrac{3}{2}.\end{array} \right.\)
Найдём нули знаменателя: \(\sqrt {6{x^2} + 7x-3} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -\dfrac{3}{2},\\x = \dfrac{1}{3}.\end{array} \right.\)
Ответ: \(\left( {\dfrac{1}{3};\dfrac{7}{{10}}} \right].\)