Помогите пожалуйста и это)
Приложения:
Ответы на вопрос
Ответил maksimcat
0
ODZ: (1-x)²>0 ⇒ x≠1
![left { {{ 3^{2x-6} textless frac{1}{27} } atop { log_{3} (1-x)^2 leq 2}} right. \ \ left { {{ 3^{2x-6} textless 3^{-3} } } atop { log_{3} (1-x)^2 leq log_{3}9 }} right. \ \ 3 textgreater 1 \ \ left { {{2x-6 textless -3} atop {(1-x)^2 leq 9}} right. \ \ left { {{2x textless 3} atop {(x-1-3)(x-1+3) leq 0}} right. \ \ left { {{x textless 1.5} atop {(x+2)(x-4) leq 0}} right. \ \ left { {{x textless 1.5} atop {-2 leq x leq 4}} right. \ \ -2 leq x textless 1.5 \ \](https://tex.z-dn.net/?f=+left+%7B+%7B%7B+3%5E%7B2x-6%7D++textless+++frac%7B1%7D%7B27%7D+%7D+atop+%7B+log_%7B3%7D+%281-x%29%5E2+leq+2%7D%7D+right.++%5C++%5C++left+%7B+%7B%7B+3%5E%7B2x-6%7D++textless++++3%5E%7B-3%7D+%7D+%7D+atop+%7B+log_%7B3%7D+%281-x%29%5E2+++leq+log_%7B3%7D9+%7D%7D+right.++%5C++%5C+3+textgreater++1+%5C++%5C++left+%7B+%7B%7B2x-6+textless++-3%7D+atop+%7B%281-x%29%5E2+leq+9%7D%7D+right.++%5C++%5C++left+%7B+%7B%7B2x+textless++3%7D+atop+%7B%28x-1-3%29%28x-1%2B3%29+leq+0%7D%7D+right.++%5C++%5C++left+%7B+%7B%7Bx+textless++1.5%7D+atop+%7B%28x%2B2%29%28x-4%29+leq+0%7D%7D+right.++%5C++%5C++left+%7B+%7B%7Bx+textless++1.5%7D+atop+%7B-2+leq+x+leq+4%7D%7D+right.++%5C++%5C+-2+leq+x+textless++1.5+%5C++%5C+ "left { {{ 3^{2x-6} textless frac{1}{27} } atop { log_{3} (1-x)^2 leq 2}} right. \ \ left { {{ 3^{2x-6} textless 3^{-3} } } atop { log_{3} (1-x)^2 leq log_{3}9 }} right. \ \ 3 textgreater 1 \ \ left { {{2x-6 textless -3} atop {(1-x)^2 leq 9}} right. \ \ left { {{2x textless 3} atop {(x-1-3)(x-1+3) leq 0}} right. \ \ left { {{x textless 1.5} atop {(x+2)(x-4) leq 0}} right. \ \ left { {{x textless 1.5} atop {-2 leq x leq 4}} right. \ \ -2 leq x textless 1.5 \ \")
с учетом одз получаем ответ:
x∈[-2;1)U(1;1.5]
с учетом одз получаем ответ:
x∈[-2;1)U(1;1.5]
Новые вопросы