Question

(2х-3) (х+8) (3х+9)≥0

помогите решииить ​

Answer 1

$\displaystyle (2x-3)(x+8)(3x+9)\geq 0\\\\ \left \{ {{(2x-3)(x+8)\geq 0} \atop {3x+9\geq 0}} \right.=\left \{ {{x \in (- \infty,-8]\cup{[\frac{3}{2},+ \infty) } \atop {x\geq -3}} \right.=x \in [\frac{3}{2},+ \infty)\\\\ \left \{ {{(2x-3)(x+8)\leq 0} \atop {3x+9\leq 0}} \right.=\left \{ {{x \in [-8,\frac{3}{2}] } \atop {x\leq -3}} \right. =x\in[-8,-3]\\\\ x\in [-8,-3] \cup [\frac{3}{2}, + \infty)$