| Номер | Условие | Ответ |
|---|---|---|
| 1. | $ (x-3)(x^2-6x+10)>0 $ | $ (3; +\infty) $ |
| 2. | $ (2-3x)(-2x^2+4x-7)>0 $ | $ \left(\dfrac{2}{3}; +\infty\right) $ |
| 3. | $ \dfrac{x^2+1}{x+4x^2}>0 $ | $ \left(-\infty; -\dfrac{1}{4}\right)\cup(0; +\infty) $ |
| 4. | $ \dfrac{2x^3+3x^2}{-x^2-2x-3}>0 $ | $ \left(-\infty; -\dfrac{3}{2}\right) $ |
| 5. | $ \dfrac{2x-1}{x-3}-\dfrac{3-x}{x+2}<0 $ | $ (-2; 3) $ |
| 6. | $ \dfrac{x-4}{x+1}+\dfrac{3+2x}{2x+1}\geqslant-5 $ | $ (-\infty; -1)\cup\left(-\dfrac{1}{2}; +\infty\right) $ |
| 7. | $ \dfrac{1}{x^2+1}-\dfrac{2}{2x+1}>0 $ | $ \left(-\infty; -\dfrac{1}{2}\right) $ |
| 8. | $ \dfrac{x}{x^2-6x+40}-\dfrac{2}{2x+1}>0 $ | $ \left(-\infty; -\dfrac{1}{2}\right)\cup\left(\dfrac{80}{13}; +\infty\right) $ |
| 9. | $ \dfrac{x+3}{2-x}-\dfrac{x+1}{x+2}<\dfrac{x^2+x-2}{4-x^2} $ | $ (-\infty; -2)\cup(2; +\infty) $ |
| 10. | $ \dfrac{8}{(x^2-6x+8)(x-1)}<\dfrac{2x-3}{x-1}-\dfrac{x-1}{x-2}-\dfrac{x}{x-4} $ | $ (-\infty; 1)\cup(2; 4) $ |
| 11. | $ \left(\dfrac{x-1}{x+2}\right)^4-\left(\dfrac{x+3}{x+1}\right)^4\leqslant0 $ | $ \left[-\dfrac{7}{5}; -1\right)\cup(-1; +\infty) $ |
| 12. | $ \dfrac{1}{3-x}+\dfrac{9}{(x-3)^2}+\dfrac{2}{x^2-9}\geqslant-\dfrac{3}{x+3} $ | $ (-3; 3)\cup(3; +\infty) $ |
| 13. | $ \dfrac{5x-7}{x^2-x}+\dfrac{1+7 x}{1-x}<\dfrac{2x+5}{x} $ | $ (-\infty; 0)\cup(1; +\infty) $ |
| 14. | $ \dfrac{1-3x^2}{x^3-1}+\dfrac{7}{1-x}<\dfrac{5x}{x^2+x+1} $ | $ (1; +\infty) $ |
| 15. | $ \dfrac{(x-2)^2}{3-x}-\dfrac{(x-3)^2}{2-x}<0 $ | $ (-\infty; 2)\cup(3; +\infty) $ |
| 16. | $ \dfrac{x}{(3-x)(x^2-x+3)}-\dfrac{2}{x^2-5x+6}<0 $ | $ (-\infty; 2)\cup(3; +\infty) $ |
| 17. | $ \dfrac{3 x+8}{x^2+x+5}+\dfrac{12}{x}<5 $ | $ (-\infty; 0)\cup(3; +\infty) $ |
| 18. | $ \dfrac{(x-3)}{(x^2+x+3)}-\dfrac{x-1}{x^2-x+2}>0 $ | $ \emptyset $ |
| 19. | $ \dfrac{2-x}{(x^2+x+3)}-\dfrac{4-x}{x^2+x-2}>0 $ | $ (-2, 1) $ |
| 20. | $ \dfrac{2-x^2}{(x^4+x^2+2)}-\dfrac{4-x^2}{x^4+4x^2-5}>0 $ | $ (-1, 1) $ |