Question

помогите умоляю очень надо

Answer 1

$(x+\frac{2}{x})(\frac{\sqrt{x^2-4x+4}-1}{\sqrt{4-x}-1})^2 \geq 3(\frac{\sqrt{x^2-4x+4}-1}{\sqrt{4-x}-1})^2\\ODZ:\\4-x\geq 0=> x \leq 4\\4-x\neq 1 => x \neq 3\\x\neq 0\\x^2-4x+4 \geq 0 => (x-2)^2 \geq 0 => x = R.\\(x+\frac{2}{x})(\frac{\sqrt{x^2-4x+4}-1}{\sqrt{4-x}-1})^2 \geq 3(\frac{\sqrt{x^2-4x+4}-1}{\sqrt{4-x}-1})^2$

$(x+\frac{2}{x})(\frac{|x-2|-1}{\sqrt{4-x}-1})^2 - 3(\frac{|x-2|-1}{\sqrt{4-x}-1})^2 \geq 0\\(x+\frac{2}{x}-3)(\frac{|x-2|-1}{\sqrt{4-x}-1})^2\geq 0\\\frac{x^2-3x+2}{x} (\frac{|x-2|-1}{\sqrt{4-x}-1})^2\geq 0\\\frac{(x-1)(x-2)(|x-2|-1)^2}{x(\sqrt{4-x}-1)^2} \geq 0\\\frac{(x-1)(x-2)(x-3)^2(x-1)^2}{x(4-x-1)} \geq 0\\\frac{(x-1)^3(x-2)(x-3)^2}{x(x-3)} \leq 0\\---(0)+++++[1]++++++[2]-----(3)++++>x\\x = (-\infty; 0) \cup [1] \cup [2; 3)$

Объединяя с ОДЗ, получаем ответ.

Ответ: $x \in (-\infty;0) \cup [1] \cup [2;3)$