Решить систему:
![\dispaystyle \left \{ {{ \frac{567-9^{-x}}{81-3^{-x}} \geq 7 \atop {log_{0.25x^2} \frac{x+12}{4} \leq 1}} \right.](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cleft%20%5C%7B%20%7B%7B%20%5Cfrac%7B567-9%5E%7B-x%7D%7D%7B81-3%5E%7B-x%7D%7D%20%20%5Cgeq%207%20%20%5Catop%20%7Blog_%7B0.25x%5E2%7D%20%5Cfrac%7Bx%2B12%7D%7B4%7D%20%5Cleq%201%7D%7D%20%5Cright.%20)
решаем неравенства
1)
![\dispaystyle \frac{576-3^{-2x}}{81-3^{-x}} \geq 7](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cfrac%7B576-3%5E%7B-2x%7D%7D%7B81-3%5E%7B-x%7D%7D%20%5Cgeq%207%20)
![\dispaystyle (\frac{1}{3})^x=y](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%28%5Cfrac%7B1%7D%7B3%7D%29%5Ex%3Dy%20)
![\dispaystyle \frac{567-y^2}{81-y} \geq 7\\ \frac{567-y^2-7*81+7y}{81-y} \geq 0\\ \frac{y(7-7y)}{81-y} \geq 0](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cfrac%7B567-y%5E2%7D%7B81-y%7D%20%5Cgeq%207%5C%5C%20%5Cfrac%7B567-y%5E2-7%2A81%2B7y%7D%7B81-y%7D%20%5Cgeq%200%5C%5C%20%5Cfrac%7By%287-7y%29%7D%7B81-y%7D%20%5Cgeq%200)
![\dispaystyle y \neq 0. y \neq 81; y=7](https://tex.z-dn.net/?f=%5Cdispaystyle%20y%20%5Cneq%200.%20y%20%5Cneq%2081%3B%20y%3D7)
+ - +
-----7----------81---
![\dispaystyle \frac{1}{3}^{x} \leq 7\\x \geq log_{1/3}7](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cfrac%7B1%7D%7B3%7D%5E%7Bx%7D%20%5Cleq%207%5C%5Cx%20%5Cgeq%20log_%7B1%2F3%7D7%20)
![\dispaystyle \frac{1}{3}^x\ \textgreater \ 81\\x\ \textless \ -4](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cfrac%7B1%7D%7B3%7D%5Ex%5C%20%5Ctextgreater%20%5C%2081%5C%5Cx%5C%20%5Ctextless%20%5C%20-4%20)
2)
![\dispaystyle log_{0.25x^2} \frac{x+12}{4} \leq 1](https://tex.z-dn.net/?f=%5Cdispaystyle%20log_%7B0.25x%5E2%7D%20%5Cfrac%7Bx%2B12%7D%7B4%7D%20%5Cleq%201%20)
1. 0.25x²>1; x∈(-oo;-2)∪(2;+oo)
![\dispaystyle \frac{x+12}{4} \leq 0.25x^2\\x+12-x^2 \leq 0\\x^2-x-12 \geq 0](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cfrac%7Bx%2B12%7D%7B4%7D%20%5Cleq%200.25x%5E2%5C%5Cx%2B12-x%5E2%20%5Cleq%200%5C%5Cx%5E2-x-12%20%5Cgeq%200%20)
x∈(-oo;-3]∪[4;+oo)
2) 0<0.25x²<1; x∈(-2;2)
![\dispaystyle \frac{x+12}{4} \geq 0.25x^2\\x+12-x^2 \geq 0\\x^2-x-12 \leq 0](https://tex.z-dn.net/?f=%5Cdispaystyle%20%20%5Cfrac%7Bx%2B12%7D%7B4%7D%20%5Cgeq%200.25x%5E2%5C%5Cx%2B12-x%5E2%20%5Cgeq%200%5C%5Cx%5E2-x-12%20%5Cleq%200%20)
x∈[-3;4] и с учетом условия x∈(-2;2)
объединяем все промежутки
---- (- 4) -------( - 3) ------( - 2) -------( - log₃7)-------(2 )----- (4 )----
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ответ : (-oo;-4)∪(-log₃7;2)∪(4;+oo)