Предмет: Алгебра,
автор: martin2020
Пожалуйста помогите решить неравенство!
Приложения:
Ответы
Автор ответа:
0
Применим метод рационализации.
![log_{0,25x^2}(frac{x+12}{4}) leq 1; ;; ; ODZ:; left { {{frac{x+12}{4} textgreater 0} atop {0,25x^2 textgreater 0,; ; 0,25x^2ne 1}} right. ; left { {{x textgreater -12} atop {xne pm 2}} right. \\(frac{x+12}{4}-0,25x^2)(0,25x^2-1) leq 0\\(frac{x+12}{4}-frac{x^2}{4})(frac{x^2}{4}-1) leq 0\\frac{x+12-x^2}{4}cdot frac{x^2-4}{4} leq 0\\frac{x^2-x-12}{4}cdot frac{(x-2)(x+2)}{4} geq 0\\(x-4)(x+3)(x-2)(x+2) geq 0\\+++(-3)---(-2)+++(2)---(4)+++\\xin (-12 ,-3, ]cup (-2,2)cup [, 4,+infty )](https://tex.z-dn.net/?f=log_%7B0%2C25x%5E2%7D%28frac%7Bx%2B12%7D%7B4%7D%29+leq+1%3B+%3B%3B+%3B+ODZ%3A%3B++left+%7B+%7B%7Bfrac%7Bx%2B12%7D%7B4%7D+textgreater++0%7D+atop+%7B0%2C25x%5E2+textgreater++0%2C%3B+%3B+0%2C25x%5E2ne+1%7D%7D+right.+%3B++left+%7B+%7B%7Bx+textgreater++-12%7D+atop+%7Bxne+pm+2%7D%7D+right.+%5C%5C%28frac%7Bx%2B12%7D%7B4%7D-0%2C25x%5E2%29%280%2C25x%5E2-1%29+leq+0%5C%5C%28frac%7Bx%2B12%7D%7B4%7D-frac%7Bx%5E2%7D%7B4%7D%29%28frac%7Bx%5E2%7D%7B4%7D-1%29+leq+0%5C%5Cfrac%7Bx%2B12-x%5E2%7D%7B4%7Dcdot+frac%7Bx%5E2-4%7D%7B4%7D+leq+0%5C%5Cfrac%7Bx%5E2-x-12%7D%7B4%7Dcdot+frac%7B%28x-2%29%28x%2B2%29%7D%7B4%7D+geq+0%5C%5C%28x-4%29%28x%2B3%29%28x-2%29%28x%2B2%29+geq+0%5C%5C%2B%2B%2B%28-3%29---%28-2%29%2B%2B%2B%282%29---%284%29%2B%2B%2B%5C%5Cxin+%28-12+%2C-3%2C+%5Dcup+%28-2%2C2%29cup+%5B%2C+4%2C%2Binfty+%29 "log_{0,25x^2}(frac{x+12}{4}) leq 1; ;; ; ODZ:; left { {{frac{x+12}{4} textgreater 0} atop {0,25x^2 textgreater 0,; ; 0,25x^2ne 1}} right. ; left { {{x textgreater -12} atop {xne pm 2}} right. \\(frac{x+12}{4}-0,25x^2)(0,25x^2-1) leq 0\\(frac{x+12}{4}-frac{x^2}{4})(frac{x^2}{4}-1) leq 0\\frac{x+12-x^2}{4}cdot frac{x^2-4}{4} leq 0\\frac{x^2-x-12}{4}cdot frac{(x-2)(x+2)}{4} geq 0\\(x-4)(x+3)(x-2)(x+2) geq 0\\+++(-3)---(-2)+++(2)---(4)+++\\xin (-12 ,-3, ]cup (-2,2)cup [, 4,+infty )")
Похожие вопросы
Предмет: Английский язык,
автор: burkovaana786
Предмет: Химия,
автор: duhailinyh72
Предмет: Математика,
автор: dianalazareva2525
Предмет: Математика,
автор: Piraya2014