Question

Решить неравенство:​

Answer 1

$\sqrt{x^2-3x+2} \geq x+3\\1. x + 3 < 0 => x^2 - 3x + 2 \geq 0 => (x-1)(x-2) \geq 0 => x \in (-\infty; 1] \cup [2; +\infty) => x \in (-\infty; -3)\\2. x + 3 \geq 0\\x^2 - 3x + 2 \geq x^2 + 6x + 9\\-9x \geq 7\\x \leq -\frac{7}{9} \\x\in [-3; -\frac{7}{9}]\\ Answer: x \in (-\infty; -\frac{7}{9}]$