Реши

\(x\ge -2,\ x\le1,\ x\ge-\frac32\)
\[ \boxed{x\in\left[-\frac32;1\right]} \]

\[ 4x^2+5x-6=(x+2)(4x-3)\ge0,\quad x\ge0 \] \[ \boxed{x\in\left[\frac34;+\infty\right)} \]

\[ 2x-4\le3x-9\Rightarrow \boxed{x\ge5} \]