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| — | math-public:metod_intervalov_izolirovannaya-tochka [2016/06/22 17:31] (текущий) – создано labreslav |
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| | =====Изолированная точка===== |
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| | ^ Номер ^ Условие ^ Ответ ^ |
| | | \\ 1. | \\ $x^8\geqslant x^4$\\ | \\ $(-\infty; -1)\cup\{0\}\cup(1; +\infty)$ | |
| | | \\ 2. | \\ $\dfrac{(x+1)^2}{x^2-x}\leqslant0$\\ | \\ $\{-1\}\cup(0; 1)$ | |
| | | \\ 3. | \\ $(x-3)^2(x^2-6x+8)\geqslant0$\\ | \\ $(-\infty; 2]\cup\{3\}\cup[4; +\infty)$ | |
| | | \\ 4. | \\ $(x^2-4x+3)(2x^2-3x-9)\leqslant0$\\ | \\ $\left[-\dfrac{3}{2}; 1\right]\cup\{3\}$ | |
| | | \\ 5. | \\ $\dfrac{(x^2-5x+4)^2}{(x-3)(x^2-9)}\leqslant0$\\ | \\ $(-\infty; -3)\cup\{1\}\cup\{4\}$ | |
| | | \\ 6. | \\ $\dfrac{x+11}{x-5}-\dfrac{x-7}{3-x}\leqslant0$\\ | \\ $\{1\}\cup(3; 5)$ | |
| | | \\ 7. | \\ $ \dfrac{x+2}{x-1}-\dfrac{x+1}{7-x}\geqslant\dfrac{x^2+x-24}{x^2-8x+7} $\\ | \\ $ (-\infty; 1)\cup\{3\}\cup(7; +\infty) $ | |
| | | \\ 8. | \\ $ \dfrac{x+3}{x+2}+\dfrac{x+1}{x-1}\geqslant\dfrac{2x}{x+3}-\dfrac{12}{(x^2+x-2)(x+3)} $\\ | \\ $ (-3; -2)\cup\{-1\}\cup(1; +\infty) $ | |
| | | \\ 9. | \\ $ \dfrac{1}{x-1}-\dfrac{1}{x+1}\geqslant\dfrac{2}{4x-5} $\\ | \\ $ (-\infty; -1)\cup\left(1; \dfrac{5}{4}\right)\cup\{2\} $ | |
| | | \\ 10. | \\ $ \dfrac{1}{x+8}-\dfrac{25}{x^2+5x-24}\leqslant1 $\\ | \\ $ (-\infty; -8)\cup\{-2\}\cup(3; +\infty) $ | |
| | | \\ 11. | \\ $ \dfrac{43 x+117}{x^2+2x-3}+\dfrac{13-4 x}{x-1}\geqslant\dfrac{48-4x}{x} $\\ | \\ $ \{-6\}\cup(-3; 0)\cup(1; +\infty) $ | |
| | | \\ 12. | \\ $ \dfrac{5}{x-2}-\dfrac{x^2+8}{x^3-8}\geqslant\dfrac{x}{x^2+2x+4} $\\ | \\ $ \{-2\}\cup(2; +\infty) $ | |
| | | \\ 13. | \\ $ \dfrac{(x^2-9x+18)^2}{(x^2-x)^2-(x^2+x)^2}\geqslant0 $\\ | \\ $ (-\infty; 0)\cup\{3\}\cup\{6\} $ | |
| | | \\ 14. | \\ $ \left(\dfrac{x}{x-2}\right)^2\leqslant\left(\dfrac{x}{x-3}\right)^2 $\\ | \\ $ \{0\}\cup\left[\dfrac{5}{2}; 3\right)\cup(3; +\infty) $ | |
| | | \\ 15. | \\ $ \dfrac{(x^2-5x+4)(x^2-6x+8)}{(x-3)(x^2-9)}\leqslant0 $\\ | \\ $ (-\infty; -3)\cup[1; 2]\cup\{4\} $ | |
| | | \\ 16. | \\ $ \dfrac{9 x+13}{2(x^2+1)}+\dfrac{1}{x+1}-\dfrac{9}{2(x-1)}\geqslant0 $\\ | \\ $ \{-2\}\cup(-1; 1)\cup[3; +\infty) $ | |
| | | \\ 17. | \\ $ \dfrac{1}{2x+4}-\dfrac{1}{x}\leqslant\dfrac{9}{4-2x} $\\ | \\ $ (-\infty; -2)\cup\{-1\}\cup(0; 2) $ | |
| | | \\ 18. | \\ $ \dfrac{15}{2-x}-\dfrac{9}{(x-2)^2}+\dfrac{25}{x-4}\leqslant-1 $\\ | \\ $ \{-1\}\cup[0; 2)\cup(2; 4) $ | |
| | | \\ 19. | \\ $ \dfrac{x+5}{x+2}-\dfrac{1}{x^2-2x+4}\leqslant\dfrac{5 x^2-8 x+18}{x^3+8} $\\ | \\ $ (-2; 0]\cup\{1\} $ | |
| | | \\ 20. | \\ $ x^2-x+3\geqslant\dfrac{16}{3 (x+2)}+\dfrac{1}{3 (1-x)} $\\ | \\ $ (-\infty; -2)\cup\{0\}\cup(1; +\infty) $ | |
