- \(\displaystyle\frac{0,5}{x-x^2-1}<0\)
- \(\displaystyle\frac{x^2-5x+6}{x^2+x+1}<0\)
- \(\displaystyle\frac{x^2+2x-3}{x^2+1}<0\)
- \(\displaystyle\frac{(x-1)(x+2)^2}{-1-x}\le0\)
- \(\displaystyle\frac{x^2+4x+4}{2x^2-x-1}>0\)
- \(\displaystyle\frac{(2-x^2)(x-3)^3}{(x+1)(x^2-3x-4)}\ge0\)
- \(\displaystyle\frac{(x+2)(x^2-2x+1)}{4+3x-x^2}\ge0\)
- \(\displaystyle\frac{x^4-3x^3+2x^2}{x^2-x-30}\ge0\)
- \(\displaystyle\frac{x^2+2}{x^2-1}<-2\)
- \(\displaystyle\frac{2(x-4)}{(x-1)(x-7)}\ge\frac{1}{x-2}\)
- \(\displaystyle\frac{7}{(x-2)(x-3)}+\frac{9}{x-3}+1\le0\)
- \(\displaystyle\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}>1\)
- \(\displaystyle\frac{x^2-6x+9}{5-4x-x^2}\ge0\)
- \(\displaystyle\frac{x+1}{x-1}\ge\frac{x+5}{x+1}\)
Ответы
- \((-\infty;+\infty)\)
- (2;3)
- (-3;1)
- \((-\infty;-1)\cup[1;+\infty)\)
- \((-\infty;-2)\cup(-2;-1/2)\cup(1;+\infty)\)
- \([-\sqrt{2};-1)\cup(-1;\sqrt{2}]\cup[3;4)\)
- \((-\infty;-2)\cup(-1;4)\)
- \((-\infty;-5)\cup[1;2]\cup[6;+\infty)\cup\{0\}\)
- \((-1;0)\cup(0;1)\)
- \((1;2)\cup(7;+\infty)\)
- \([-5;1]\cup(2;3)\)
- \((-\infty;-7)\cup(-4;-2)\)
- \((-1;1]\cup(4;6]\)
- \((-5;-1)\cup\{3\}\)
- \((-\infty;-1)\cup(1;3]\)