Question

Решите неравенства.
√x^{2} -6X+9≤5
|5-3X|≥5

Answer 1

1)

$\sqrt{x^2}-6x+9 \leq 5\\ \to\\|x|-6x \leq -4, ~x\geq 0\\|x|-6x \leq -4, ~x < 0\\\\x \geq \frac{4}{5}, ~x\geq 0\\x \geq \frac{4}{7}, ~x<0$

x ∈ ∅

Ответ: x ∈ $[\frac{4}{5}, + \infty]$

2)

$|5-3x|>5\\\to\\5-3x>5,~~ \to ~~5-3x\geq0\\-(5-3x)>5,~~ \to~~5-3x<0\\\\x < 0, x \leq \frac{5}{3}\\x > \frac{10}{3}, x > \frac{5}{3}$

Ответ: x ∈ $(- \infty, 0)$ ∪ $(\frac{10}{3}, + \infty)$