=====Множитель постоянного знака===== ^ 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) $ |