Question

СРОЧНО ДАЮ 50 БАЛЛООВ

Answer 1

$\displaystyle \tt x^{2}-5x+6\geq0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=25-24=1\\\\x_{1,2}=\frac{-bб\sqrt{D}}{2a}\\\\x_{1}=3\\\\x_{2}=2\\\\\\x^{2}-5x+6\geq0\\\\(x-3)(x-2)\geq0\\\\x\in(-\infty;2]\cup[3;\infty)$