| Номер | Условие | Ответ |
|---|---|---|
| 1. | $(x-3)(2-x)\leqslant0$ | $(-\infty; 2]\cup[3; +\infty)$ |
| 2. | $(x+3)(4-x)(5-x)(1-x)<0$ | $(-\infty; -3)\cup(1; 4)\cup(5; +\infty)$ |
| 3. | $(-x^2-7)(5x-7)>0$ | $\left(-\infty; \dfrac{7}{5}\right)$ |
| 4. | $\dfrac{x-6}{7-x}\geqslant0$ | $[6; 7)$ |
| 5. | $\dfrac{6x-x^2}{x+1}>0$ | $(-\infty; -1)\cup(0; 6)$ |
| 6. | $\dfrac{7x^5-x^4}{4-x^2}<0$ | $(-2; 0)\cup\left(0; \dfrac{1}{7}\right)\cup(2; +\infty)$ |
| 7. | $(3-x-2x^2)(x^2-5x+4)\leqslant0$ | $\left(-\infty; -\dfrac{3}{2}\right]\cup\{1\}\cup[4; +\infty)$ |
| 8. | $\dfrac{x-1}{x-2}-\dfrac{3x-1}{x-4}<0$ | $\left(-\infty; \dfrac{1-\sqrt{5}}{2}\right)\cup\left(\dfrac{1+\sqrt{5}}{2}; 2\right)\cup(4; +\infty)$ |
| 9. | $\dfrac{2-x}{3-x}-\dfrac{7x+5}{x+2}<\dfrac{4x+17}{x^2-x-6}$ | $(-\infty; -2)\cup(3; +\infty)$ |
| 10. | $(x-1)^2-(x^2-3x-2)^2<0$ | $(-\infty; -1)\cup(2-\sqrt{5}; 3)\cup(2+\sqrt{5}; +\infty)$ |
| 11. | $\dfrac{1}{x-x^2+2}-\dfrac{2}{6-7x+x^2}\leqslant0$ | $(-\infty;-1)\cup\left[\dfrac{9-\sqrt{57}}{6};1\right)\cup\left(2; \dfrac{9+\sqrt{57}}{6}\right]\cup(6;+\infty)$ |
| 12. | $\dfrac{x}{5x-3x^2-2}+\dfrac{x}{1+2x+3x^2}<0$ | $(-\infty; 0)\cup\left(\dfrac{1}{7}; \dfrac{2}{3}\right)\cup(1; +\infty)$ |
| 13. | $\dfrac{3}{x}-\dfrac{x}{x^2+2x-3}<\dfrac{2}{x-1}$ | $(-3, 0)\cup(1, +\infty)$ |
| 14. | $\dfrac{9-x}{1+2x}-\dfrac{x}{2-5x}>0$ | $\left(-\dfrac{1}{2}; 8-\sqrt{58}\right)\cup\left(\dfrac{2}{5}; 8+\sqrt{58}\right)$ |
| 15. | $\dfrac{1}{x^2+25}-\dfrac{1}{x-5}+\dfrac{1}{x+5}>0$ | $(-5; 5)$ |
| 16. | $\dfrac{(x^2-3x)^2-(3x^2+6x)^2}{(16x^2+24x+9)^2}\geqslant0$ | $\left[-\dfrac{9}{2}, -\dfrac{3}{4}\right)\cup\{0\}$ |
| 17. | $\dfrac{(x-37-x^2)(x^2-13x+30)}{(10-x)(1-x)}\geqslant0$ | $(1;3]$ |
| 18. | $\dfrac{1}{2x^2-x-1}-\dfrac{1}{x^2-5x+4}<\dfrac{1}{5-2x^2-3x}$ | $\left(-\dfrac{5}{2}; -\dfrac{29}{22}\right)\cup\left(-\dfrac{1}{2}; 1\right)\cup(4; +\infty)$ |
| 19. | $\dfrac{(4-x)^2(2-x)^3(21-x)}{(7-x)(-x-1)(-2x+5)}\geqslant0$ | $(-\infty; -1)\cup\left[2; \dfrac{5}{2}\right)\cup\{4\}\cup(7; 21]$ |
| 20. | $\dfrac{(x-4)^2(2-5x)^9(2-3x)^8}{(12+x)^4(x-1)^3(-2x+9)^8}\geqslant0$ | $\left[\dfrac{2}{5}; 1\right)\cup\{4\}$ |