- \((x-1)(3-x)(x-2)^2>0\)
- \(\displaystyle\frac{6x-5}{4x+1}<0\)
- \(\displaystyle\frac{2x-3}{3x-7}>0\)
- \(\displaystyle\frac{3}{x-2}<1\)
- \(\displaystyle\frac{4x+3}{2x-5}<6\)
- \(\displaystyle\frac{x}{x-5}>\frac{1}{2}\)
- \(\displaystyle\frac{2x+3}{x^2+x-12}\le\frac{1}{2}\)
- \(\displaystyle\frac{4}{1+x}+\frac{2}{1-x}<1\)
- \(2+\displaystyle\frac{3}{x+1}>\frac{2}{x}\)
- \(\displaystyle\frac{2(x-3)}{x(x-6)}\le\frac{1}{x-1}\)
- \(\displaystyle\frac{1}{x-2}+\frac{1}{x-1}>\frac{1}{x}\)
- \(\displaystyle\frac{2x}{x^2-9}\le\frac{1}{x+2}\)
- \(\displaystyle\frac{x^2-6x+9}{5-4x-x^2}\ge0\)
- \(\displaystyle\frac{x^2-1}{x^2+x+1}<1\)
- \(\displaystyle\frac{(2-x^2)(x-3)^3}{(x+1)(x^2-3x-4)}\ge0\)
Ответы
- \((1;2)\cup(2;3)\)
- \((-1/4;5/6)\)
- \((-\infty;3/2)\cup(7/3;+\infty)\)
- \((-\infty;2)\cup(5;+\infty)\)
- \((-\infty;5/2)\cup(33/8;+\infty)\)
- \((-\infty;-5)\cup(5;+\infty)\)
- \((-\infty;-4)\cup[-3;3)\cup[6;+\infty)\)
- \((-\infty;-1)\cup(1;+\infty)\)
- \((-\infty;-2)\cup(-1;0)\cup(1/2;+\infty)\)
- \((-\infty;0)\cup(1;6)\)
- \((-\sqrt{2};0)\cup(1;\sqrt{2})\cup(2;+\infty)\)
- \((-\infty;-3)\cup(-2;3)\)
- \((-5;1)\cup\{3\}\)
- \((-2;+\infty)\)
- \([-\sqrt{2};-1)\cup(-1;\sqrt{2}]\cup[3;4)\)