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| math-public:metod_intervalov_nelzya-domnozhat-i-sokrashchat [2016/06/22 17:32] – создано labreslav | math-public:metod_intervalov_nelzya-domnozhat-i-sokrashchat [2016/08/18 17:58] (текущий) – labreslav |
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| | =====Нельзя домножать и сокращать===== |
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| | ^ Номер ^ Условие ^ Ответ ^ |
| | | \\ 1. | \\ $ x^4<x^2 $\\ | \\ $ (-1; 0)\cup(0; 1) $ | |
| | | \\ 2. | \\ $ x^2>x $\\ | \\ $ (-\infty; 0)\cup(1; +\infty) $ | |
| | | \\ 3. | \\ $ x-\dfrac{1}{x}\geqslant\dfrac{1}{2} $\\ | \\ $ \left[\dfrac{1-\sqrt{17}}{4}; 0\right)\cup\left[\dfrac{1+\sqrt{17}}{4}; +\infty\right) $ | |
| | | \\ 4. | \\ $ \dfrac{4x-3}{5x-7}>3 $\\ | \\ $ \left(\dfrac{7}{5}; \dfrac{18}{11}\right) $ | |
| | | \\ 5. | \\ $ \dfrac{4x^2-4x+1}{5x-7}\geqslant0 $\\ | \\ $ \left\{\dfrac{1}{2}\right\}\cup\left(\dfrac{7}{5}; +\infty\right) $ | |
| | | \\ 6. | \\ $ x-4\geqslant\dfrac{1}{7-x} $\\ | \\ $ \left[\dfrac{11-\sqrt{5}}{2}; \dfrac{11+\sqrt{5}}{2}\right]\cup(7; +\infty) $ | |
| | | \\ 7. | \\ $ \dfrac{2x-5}{4x-3}\geqslant\dfrac{x-1}{x-7} $\\ | \\ $ \left[-8; \dfrac{3}{4}\right)\cup[2; 7) $ | |
| | | \\ 8. | \\ $ \dfrac{3x+13}{3x^2-5x-12}<1 $\\ | \\ $ \left(-\infty; \dfrac{4-\sqrt{91}}{3}\right)\cup\left(-\dfrac{4}{3}; 3\right)\cup\left(\dfrac{4+\sqrt{91}}{3}; +\infty\right) $ | |
| | | \\ 9. | \\ $ \dfrac{x^2+6x-7}{x^2-6x+5}<\dfrac{x}{x-5} $\\ | \\ $ (-\infty; 1)\cup(1; 5) $ | |
| | | \\ 10. | \\ $ \dfrac{x^2+6x-7}{x^2-6x+5}\geqslant\dfrac{x+7}{x-4} $\\ | \\ $ [-7; 1)\cup(1; 4)\cup(5; +\infty) $ | |
| | | \\ 11. | \\ $ \dfrac{x+3}{x^2-10x+25}\geqslant\dfrac{x^2+6x+9}{x-5} $\\ | \\ $ (-\infty; 1-\sqrt{17}]\cup[-3; 5)\cup(5; 1+\sqrt{17}] $ | |
| | | \\ 12. | \\ $ \dfrac{3}{x-1}-\dfrac{1}{x^2-2x+1}\leqslant1 $\\ | \\ $ (-\infty; 1)\cup\left(1; \dfrac{5-\sqrt{5}}{2}\right]\cup\left[\dfrac{5+\sqrt{5}}{2}; +\infty\right) $ | |
| | | \\ 13. | \\ $ \dfrac{2x(3+4x)}{2+x}\leqslant\dfrac{(x+1)(3+4x)}{4-x} $\\ | \\ $ (-\infty; -2)\cup\left[-\dfrac{3}{4}; \dfrac{2}{3}\right]\cup[1; 4) $ | |
| | | \\ 14. | \\ $ \dfrac{(x^2+8x+16)(3+x)}{2-x}\leqslant\dfrac{(x^2+8x+16)(2x-1)}{2-x} $\\ | \\ $ \{-4\}\cup(2; 4] $ | |
| | | \\ 15. | \\ $ \dfrac{(x^2+8x+16)(x^2-4x+7)}{x+3}\leqslant\dfrac{(x^2+8x+16)(2x^2-x-7)}{x+3} $\\ | \\ $ \left[\dfrac{-3-\sqrt{65}}{2}; -3\right)\cup\left[\dfrac{\sqrt{65}-3}{2}; +\infty\right) $ | |
| | | \\ 16. | \\ $ \dfrac{(2x^2+x-1)(3+x)}{5-x}\leqslant\dfrac{(x^2+6x+9)(2x-1)}{x+1} $\\ | \\ $ (-\infty; -3]\cup[-\sqrt{7}; -1)\cup\left[\dfrac{1}{2}; \sqrt{7}\right]\cup(5; +\infty) $ | |
| | | \\ 17. | \\ $ \dfrac{(2x^2+3x+1)(1-4x)}{x+5}>\dfrac{(2x^2+3x+1)(2x-5)}{x-1} $\\ | \\ $ (-5; -2)\cup\left(-1; -\dfrac{1}{2}\right)\cup(1; 2) $ | |
| | | \\ 18. | \\ $ \dfrac{(9x^2+6x+1)(1+x)}{x+4}\leqslant\dfrac{(9x^2+6x+1)(x-2)}{2-3x} $\\ | \\ $ (-4, -2]\cup\left\{-\dfrac{1}{3}\right\}\cup\left(\dfrac{2}{3}; \dfrac{5}{4}\right] $ | |
| | | \\ 19. | \\ $ \dfrac{7-x}{x^2-6x+5}-\dfrac{7-x}{2x^2-x-1}\leqslant\dfrac{7-x}{2x^2+x-3} $\\ | \\ $ \left(-\infty; -\dfrac{3}{2}\right)\cup\left[-\dfrac{23}{24}; -\dfrac{1}{2}\right)\cup(1; 5)\cup[7; +\infty) $ | |
| | | \\ 20. | \\ $ \dfrac{6+5x-x^2}{x^2-6x+5}-\dfrac{(6-x)(x+2)}{x^2-2x+1}>\dfrac{x-6}{2x^2+x-3} $\\ | \\ $ \left(-\infty; -\dfrac{3}{2}\right)\cup(5; 6) $ | |
| | | \\ 21. | \\ $ \dfrac{1}{x^2-6x+5}-\dfrac{1}{x^2-2x+1}\leqslant\dfrac{2}{2x^2+x-3}-\dfrac{2}{2x^2+3x-5} $\\ | \\ $ \left(-\infty; \dfrac{-11-\sqrt{91}}{3}\right]\cup\left(-\dfrac{5}{2}; -\dfrac{3}{2}\right)\cup\left[\dfrac{\sqrt{91}-11}{3}; 1\right)\cup(1; 5) $ | |
| | | \\ 22. | \\ $ \dfrac{1}{x^2-3x+2}-\dfrac{1}{x^2-7x+12}<\dfrac{1}{x^2-4x+3}-\dfrac{1}{x^2-6x+8} $\\ | \\ $ (1; 2)\cup\left(\dfrac{5}{2}; 3\right)\cup(4; +\infty) $ | |
