Предмет: Математика,
автор: andrushalolo
Решите неравенство:
2/(7^x - 7) ≥ 5/(7^x - 4)
Ответы
Автор ответа:
48
ОДЗ
![7^x \neq 7 \\ \\ x \neq 1 \\ \\ 7^x \neq 4 \\ \\ x \neq log_74 7^x \neq 7 \\ \\ x \neq 1 \\ \\ 7^x \neq 4 \\ \\ x \neq log_74](https://tex.z-dn.net/?f=7%5Ex+%5Cneq+7+%5C%5C++%5C%5C+x+%5Cneq+1+%5C%5C++%5C%5C+7%5Ex+%5Cneq+4+%5C%5C++%5C%5C+x+%5Cneq+log_74)
![\frac{2}{7^x - 7} \geq \frac{5}{7^x - 4} \\ \\ \frac{2}{7^x - 7} - \frac{5}{7^x - 4} \geq 0 \\ \\ \frac{2*(7^x-4)-5(7^x-7)}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{2*7^x-8-5*7^x+35}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{-3*7^x+27}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{-3*(7^x-9)}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{7^x-9}{(7^x - 7)(7^x-4)} \leq 0 \frac{2}{7^x - 7} \geq \frac{5}{7^x - 4} \\ \\ \frac{2}{7^x - 7} - \frac{5}{7^x - 4} \geq 0 \\ \\ \frac{2*(7^x-4)-5(7^x-7)}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{2*7^x-8-5*7^x+35}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{-3*7^x+27}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{-3*(7^x-9)}{(7^x - 7)(7^x-4)} \geq 0 \\ \\ \frac{7^x-9}{(7^x - 7)(7^x-4)} \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7B7%5Ex+-+7%7D++%5Cgeq+++%5Cfrac%7B5%7D%7B7%5Ex+-+4%7D+%5C%5C++%5C%5C++%5Cfrac%7B2%7D%7B7%5Ex+-+7%7D+-++%5Cfrac%7B5%7D%7B7%5Ex+-+4%7D++%5Cgeq+0+%5C%5C++%5C%5C+%5Cfrac%7B2%2A%287%5Ex-4%29-5%287%5Ex-7%29%7D%7B%287%5Ex+-+7%29%287%5Ex-4%29%7D++%5Cgeq+0+%5C%5C++%5C%5C++%5Cfrac%7B2%2A7%5Ex-8-5%2A7%5Ex%2B35%7D%7B%287%5Ex+-+7%29%287%5Ex-4%29%7D++%5Cgeq+0+%5C%5C++%5C%5C+%5Cfrac%7B-3%2A7%5Ex%2B27%7D%7B%287%5Ex+-+7%29%287%5Ex-4%29%7D++%5Cgeq+0+%5C%5C++%5C%5C++%5Cfrac%7B-3%2A%287%5Ex-9%29%7D%7B%287%5Ex+-+7%29%287%5Ex-4%29%7D++%5Cgeq+0+%5C%5C++%5C%5C+%5Cfrac%7B7%5Ex-9%7D%7B%287%5Ex+-+7%29%287%5Ex-4%29%7D+%5Cleq+0)
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______________log₇4_________1__________log₇9______
x∈(-∞; log₇4)∪(1; log₇9]
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- + - +
______________log₇4_________1__________log₇9______
x∈(-∞; log₇4)∪(1; log₇9]
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