помогите решить пж 20 баллов. Created by dfgrtyhdg2112433. algebra-ru

Объяснение:как же, я увидел)где?6*log(x+4) = log(x+4)^6 <= log6 (6^(12))(x+4)^6 <= 6^12|x+4| <= 6^2-36 <= x+4 <= 36-40 <= x <= 32а почему нельзя сократить на 6 ? т.к. для x < -4 и выражение (x-3)<0 и (x+4)<0 и под логарифмом они (эти выражения) находиться не могут... lg(x^2) = 2*lg(|x|)Участник ЗнанийУчастник ЗнанийОтвет: x ∈ [-40; -4) ∪ (3;32].Объяснение:Новые вопросы в АлгебраПодайте у вигляді суми квадратів двох виразів многочлен-25х2-24xу +9y2;в классе учатся 28 человек. в один день каждый принес по три фломастера красный,зеленый,синий.Могут ли ученики класса обменяться фломастерами так, что…у каждого оказалось 3 фломастера одинакого цветаvery very срочно!помогите пожалуйста.Постройте график функции y = (3x^2 + 4x)/x - (x^2 - 1)/(x + 1)даю 20 баллов за простенькое задание Нам дано число от 0 до бесконечности. Придумать формулу, которая будет преобразовывать это число от 30 до 12 соответственно. То есть бесконечность фор…мула будет преобразовывать в число 12. А ноль будет преобразовывать в 30. Например число между условной бесконечностью и нулем будет число 21. Придумать универсальную формулу, чтобы можна было подставлять числа.Пожалуйста помогитеab умножить на 2 сколько будет? срочно, даю 30 балловмолю про допомогу)) якщо мало ось мій інст))) Помогите пожалуйсто как ето решытьПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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помогите решить пж 20 баллов...

dfgrtyhdg2112433 @dfgrtyhdg2112433

July 2022 1 5 Report

помогите решить пж 20 баллов

Answers & Comments

NNNLLL54

Verified answer

Объяснение:

$5log_6(x^2+x-12)\leq 12+log_6\frac{(x-3)^5}{x+4}\\\\ODZ:\; \; \left \{ {{x^2+x-12>0} \atop {\frac{(x-3)^5}{x+4}>0} \right.\; \; \left \{ {{(x-3)(x+4)>0} \atop {\frac{(x-3)^5}{x+4}>0}} \right.\; \; \left \{ {{x3} \atop {x3}} \right.\\\\x\in (-\infty ,-4)\cup (3,+\infty )\\\\5\, log_6((x-3)(x+4))\leq 12+log_6|x-3|^5-log_6|x+4|\\\\5\cdot log_6|x-3|+5\cdot log_6|x+4|\leq 12+5\cdot log_6|x-3|-log_6|x+4|\\\\6\cdot log_6|x+4|\leq 12\\\\log_6|x+4|\leq 2\\\\log_6|x+4|\leq log_66^2\; \; \Rightarrow \; \; t.k.\; \; a=6>1\; ,\; to\; \; |x+4|\leq 36\; ,\\\\-36\leq x+4\leq 36\; \; ,\; \; -40\leq x\leq 32\\\\\left \{ {{x\in (-\infty ,-4)\cup (3,+\infty )} \atop {-40\leq x\leq 32}} \right.\; \; \; \Rightarrow \; \; x\in [-40,-4)\cup (3,32\, ]\\\\Otvet:\; \; x\in [-40,-4)\cup (3,32\, ]\; .$

$P.S.\; \; log_{a}(xy)=log_{a}|x|+log_{a}|y|\; \; ,\; esli\; \; xy>0\; ,\\\\log_{a}\frac{x}{y}=log_{a}|x|-log_{a}|y|\; \; ,\; esli\; \; \frac{x}{y}>0\\\\xy>0\; \; i\; \; \frac{x}{y}>0\; \; pri\; \; \{x>0\; i\; y>0\}\; \; ili\; \; \; \{x$

как же, я увидел)
где?
6*log(x+4) = log(x+4)^6 <= log6 (6^(12))
(x+4)^6 <= 6^12
|x+4| <= 6^2
-36 <= x+4 <= 36
-40 <= x <= 32
а почему нельзя сократить на 6 ?
т.к. для x < -4 и выражение (x-3)<0 и (x+4)<0 и под логарифмом они (эти выражения) находиться не могут... lg(x^2) = 2*lg(|x|)