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