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| math-public:metod_intervalov_pravij_znak_minus [2016/06/22 17:29] – создано labreslav | math-public:metod_intervalov_pravij_znak_minus [2016/08/18 19:12] (текущий) – labreslav |
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| | =====Правый знак минус===== |
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| | ^ 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. | \\ $\dfrac{(x-4)^2(2-5x)^9(2-3x)^8}{(12+x)^4(x-1)^3(-2x+9)^8}\geqslant0$\\ | \\ $\left[\dfrac{2}{5}; 1\right)\cup\{4\}$ | |
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