Question

ПОЖАЛУЙСТА ОЧЕНЬ СРОЧНО! РЕШИТЕ ДВЕ НЕРАВЕНСТВ.ЗАДАЧКИ:29;30. Не отбрасывая знаменатель.​

Answer 1

$1)\ \ \dfrac{x^2-5x+12}{x^2-4x+5}>3\ \ ,\ \ \dfrac{x^2-5x+12-3x^2+12x-15}{x^2-4x+5}>0\ ,\\\\\\\dfrac{-2x^2+7x-3}{x^2-4x+5}>0\ \ ,\ \ \dfrac{2x^2-7x+3}{x^2-4x+5}<0\ \ ,\ \ \dfrac{2(x-3)(x-\frac{1}{2})}{x^2-4x+5}<0\ ,\\\\\\x^2-4x+5=0\ \ ,\ \ D=16-20=-4<0\ \ \to \\\\x^2-4x+5>0\ \ pri\ \ x\in R\\\\(x-3)(x-\dfrac{1}{2})<0\\\\znaki:\ \ +++(\dfrac{1}{2})---(3)+++\\\\x\in \Big(\, \dfrac{1}{2}\ ;\ 3\, \Big)$

$2)\ \ \dfrac{2}{x^2-3x-4}\geq \dfrac{3}{x^2+x-6}\ \ ,\ \ \dfrac{2x^2+2x-12-3x^2+9x+12}{(x+1)(x-4)(x+3)(x-2)}\geq 0\ ,\\\\\\\dfrac{-x^2+11x}{(x+1)(x+3)(x-2)(x-4)}\geq 0\ \ ,\ \ \ \dfrac{-x(x-11)}{(x+1)(x+3)(x-2)(x-4)}\geq 0\ ,\\\\\\\dfrac{x(x-11)}{(x+1)(x+3)(x-2)(x-4)}\leq 0\\\\\\znaki:\ +++(-3)---(-1)+++[\, 0\, ]---(2)+++(4)---[\, 11\, ]+++\\\\x\in (-3\, ;-1\, )\cup [\ 0\, ;\, 2)\cup (\ 4\, ;\ 11\ ]$