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

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

x ∈ ∅

Ответ: x ∈ $[\frac{4}{5}, + \infty]$

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}\\" alt="|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}\\" align="absmiddle" class="latex-formula">

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