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| — | math-public:metod_intervalov_ocenivat-korni [2016/06/22 17:30] (текущий) – создано labreslav |
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| | =====Оценивать корни===== |
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| | ^ Номер ^ Условие ^ Ответ ^ |
| | | \\ 1. | \\ $ (x^2-3x-5)(x-4)>0 $\\ | \\ $ \left(\dfrac{3-\sqrt{29}}{2}; 4\right)\cup\left(\dfrac{3+\sqrt{29}}{2}; +\infty\right) $ | |
| | | \\ 2. | \\ $ (x^2-5x+2)(2x-1)\leqslant0 $\\ | \\ $ \left(-\infty; \dfrac{5-\sqrt{17}}{2}\right]\cup\left[\dfrac{1}{2}; \dfrac{5+\sqrt{17}}{2}\right] $ | |
| | | \\ 3. | \\ $ \dfrac{2x^2-7x+1}{4x^2-x-1}\leqslant0 $\\ | \\ $ \left(\dfrac{1-\sqrt{17}}{8}; \dfrac{7-\sqrt{41}}{4}\right]\cup\left(\dfrac{1+\sqrt{17}}{8}; \dfrac{7+\sqrt{41}}{4}\right] $ | |
| | | \\ 4. | \\ $ \dfrac{x+1}{3-2x}+\dfrac{1-7x}{2-x}\geqslant0 $\\ | \\ $ \left(-\infty; \dfrac{11-2\sqrt{14}}{13}\right]\cup\left[\dfrac{11+2\sqrt{14}}{13}; \dfrac{3}{2}\right)\cup(2; +\infty) $ | |
| | | \\ 5. | \\ $ \dfrac{2}{3-4x}+\dfrac{1}{2-3x}\geqslant\dfrac{3}{x-1} $\\ | \\ $ \left(-\infty; \dfrac{2}{3}\right)\cup\left[\dfrac{34-\sqrt{6}}{46}; \dfrac{3}{4}\right)\cup\left[\dfrac{34+\sqrt{6}}{46}; 1\right) $ | |
| | | \\ 6. | \\ $ \dfrac{x-1}{3-x}-\dfrac{x}{2x-1}<\dfrac{1-3x}{x+1} $\\ | \\ $ \left(-1; \dfrac{17-\sqrt{241}}{6}\right)\cup\left(\dfrac{1}{2}; 1\right)\cup\left(3; \dfrac{17+\sqrt{241}}{6}\right) $ | |
| | | \\ 7. | \\ $ (5x-3)(3x-2)(7x-5)(2x-1)>0 $\\ | \\ $ \left(-\infty; \dfrac{1}{2}\right)\cup\left(\dfrac{3}{5}; \dfrac{2}{3}\right)\cup\left(\dfrac{5}{7}; +\infty\right) $ | |
| | | \\ 8. | \\ $ \dfrac{6 x^2+7 x-55}{(3 x-7) (4 x+15)}\geqslant0 $\\ | \\ $ \left(-\infty; -\dfrac{15}{4}\right)\cup\left[-\dfrac{11}{3}; \dfrac{7}{3}\right)\cup\left[\dfrac{5}{2}; +\infty\right) $ | |
| | | \\ 9. | \\ $ \dfrac{1}{2x+7}+\dfrac{1}{2x-7}<\dfrac{3}{x-2}+\dfrac{3}{x+2} $\\ | \\ $ \left(-\sqrt{\dfrac{139}{10}}; -\dfrac{7}{2}\right)\cup(-2; 0)\cup\left(2; \dfrac{7}{2}\right)\cup\left(\sqrt{\dfrac{139}{10}}; +\infty\right) $ | |
| | | \\ 10. | \\ $ \dfrac{2}{3x+5}-\dfrac{2}{3x-5}\geqslant\dfrac{1}{x-1}-\dfrac{1}{x+1} $\\ | \\ $ \left(-\dfrac{5}{3}; -\sqrt{\dfrac{35}{19}}\right]\cup(-1; 1)\cup\left[\sqrt{\dfrac{35}{19}}; \dfrac{5}{3}\right) $ | |
| | | \\ 11. | \\ $ \dfrac{5}{x-4}-\dfrac{256}{x^3-64}>\dfrac{x}{x^2+4x+16} $\\ | \\ $ (-3-\sqrt{53}; 4)\cup(\sqrt{53}-3; +\infty) $ | |
| | | \\ 12. | \\ $ \left(\dfrac{x-23}{x+5}+\dfrac{2}{x}+1\right)\cdot\dfrac{x}{4x^2-3}\geqslant0 $\\ | \\ $ \left(-5; -\dfrac{\sqrt{3}}{2}\right)\cup\left[4-\sqrt{11}; \dfrac{\sqrt{3}}{2}\right)\cup[4+\sqrt{11}; +\infty) $ | |
| | | \\ 13. | \\ $ \dfrac{1}{2x-1}+\dfrac{1}{x+2}>\dfrac{1}{3x+7} $\\ | \\ $ \left(\dfrac{-21-3\sqrt{21}}{14}; -\dfrac{7}{3}\right)\cup\left(-2; \dfrac{3\sqrt{21}-21}{14}\right)\cup\left(\dfrac{1}{2}; +\infty\right) $ | |
| | | \\ 14. | \\ $ \left(\dfrac{x^2+3x+2}{x-2}\right)^2\geqslant\left(\dfrac{x^2-1}{x-3}\right)^2 $\\ | \\ $ \{-1\}\cup[1-\sqrt{3}; 2)\cup(2; 1+\sqrt{3}]\cup[4; +\infty) $ | |
| | | \\ 15. | \\ $ \dfrac{x^2+3x+1}{3x-5}\leqslant\dfrac{3x-5}{x^2+3x+1} $\\ | \\ $ (-\infty; -3-\sqrt{13}]\cup\left(\dfrac{-3-\sqrt{5}}{2}; \dfrac{\sqrt{5}-3}{2}\right)\cup\left[\sqrt{13}-3; \dfrac{5}{3}\right) $ | |
| | | \\ 16. | \\ $ \left(\dfrac{2x^2+4x+1}{3x-2}\right)^2>\left(\dfrac{3x-2}{2x^2+4x+1}\right)^2 $\\ | \\ $ \left(-\infty; \dfrac{-7-\sqrt{57}}{4}\right)\cup\left(\dfrac{\sqrt{57}-7}{4}; \dfrac{2}{3}\right)\cup\left(\dfrac{2}{3}; +\infty\right) $ | |
| | | \\ 17. | \\ $ \dfrac{2x+1}{3x-2}>\dfrac{3x-2}{2x+1}+2 $\\ | \\ $ \left(-\dfrac{1}{2}; \dfrac{9-7\sqrt{2}}{17}\right)\cup\left(\dfrac{2}{3}; \dfrac{9+7\sqrt{2}}{17}\right) $ | |
| | | \\ 18. | \\ $ \dfrac{x^2+x-1}{x^2+x-2}-\dfrac{x^2+x-2}{x^2+x-3}<0 $\\ | \\ $ \left(-\infty; \dfrac{-1-\sqrt{13}}{2}\right)\cup(-2; 1)\cup\left(\dfrac{\sqrt{13}-1}{2}; +\infty\right) $ | |
| | | \\ 19. | \\ $ 4x^2+2x-\dfrac{16x^4+16x^3+2x-2}{4x^2+2x-3}>0 $\\ | \\ $ \left(\dfrac{-1-\sqrt{2}}{2}; \dfrac{-1-\sqrt{13}}{4}\right)\cup\left(\dfrac{\sqrt{2}-1}{2}; \dfrac{\sqrt{13}-1}{4}\right)$ | |
| | | \\ 20. | \\ $ \dfrac{x^2+2x+1}{x^2+4x+4}+3\dfrac{x^2-1}{x^2-4}+2\dfrac{x^2-2x+1}{x^2-4x+4}\leqslant0 $\\ | \\ $ \left[\dfrac{-1-\sqrt{73}}{6}; -\sqrt{2}\right]\cup\left[\dfrac{\sqrt{73}-1}{6}; \sqrt{2}\right] $ | |
