\dfrac{x^2(2x-7)}{(x+3)^3} \lt 0 x \in (-3; 0)\cup \left(0; 3 \cfrac 12\right) x \in (- \infty; -3) \cup \left(0 ;3 \cfrac 12\right) x \in \left(- \infty; 3 \cfrac 12\right) x \in \left(3 \cfrac 12; +\infty\right)

Задание

Реши неравенство

\(\dfrac{x^2(2x-7)}{(x+3)^3} \lt 0\)

\(x \in (-3; 0)\cup \left(0; 3 \cfrac 12\right)\)
\(x \in (- \infty; -3) \cup \left(0 ;3 \cfrac 12\right)\)
\(x \in \left(- \infty; 3 \cfrac 12\right)\)
\( x \in \left(3 \cfrac 12; +\infty\right)\)