\cfrac{-15}{-2x^2+5x+3} \leqslant 0 x \in \left(-\cfrac 12; 3\right) x \in \left[-\cfrac 12; 3\right] x \in \left(- \infty; -\cfrac 12\right]\cup [3; +\infty) x \in (3; +\infty)

Задание

Реши неравенство

\(\cfrac{-15}{-2x^2+5x+3} \leqslant 0\)

\(x \in \left(-\cfrac 12; 3\right)\)
\(x \in \left[-\cfrac 12; 3\right]\)
\(x \in \left(- \infty; -\cfrac 12\right]\cup [3; +\infty)\)
\( x \in (3; +\infty)\)