ЛОГАРИФМ! ПОМОГИТЕ ПОЖАЛУЙСТА!. Created by bapataidamir2004. algebra-ru

Объяснение: ОДЗ:(x+2)(x-2)>0-∞__+__-2__-__2__+__+∞x∈(-∞;-2)U(2;+∞).(x+3)(x-3)<0-∞__+__-3__-__3__+__+∞x∈(-3;3). ⇒Ответ: x∈(-3;-2)U(2;3).

ЛОГАРИФМ! ПОМОГИТЕ ПОЖАЛУЙСТА!...

sangers1959

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Объяснение:

$f(x)=log_{1,5}(x^2-4)+log_{1,5}(9-x^2).$

ОДЗ:

$\left \{ {{x^2-4>0} \atop {9-x^2>0}} \right.\ \ \ \ \left \{ {{x^2-2^2>0} \atop {x^2-3^20} \atop {(x+3)(x-3)>0}} \right..$

(x+2)(x-2)>0

-∞__+__-2__-__2__+__+∞

x∈(-∞;-2)U(2;+∞).

(x+3)(x-3)<0

-∞__+__-3__-__3__+__+∞

x∈(-3;3). ⇒

Ответ: x∈(-3;-2)U(2;3).

1 votes Thanks 1

NNNLLL54

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Ответ:

$f(x)=log_{1,5}\, (x^2-4)+log_3\, (9-x^2)\\\\\\ODZ:\ \left\{\begin{array}{l}x^2-4>0\\9-x^2>0\end{array}\right\ \ \left\{\begin{array}{l}(x-2)(x+2)>0\\(3-x)(3+x)>0\end{array}\right\ \ \left\{\begin{array}{l}(x-2)(x+2)>0\\(x-3)(x+3)$

${}\ \ \ \left\{\begin{array}{l}x\in (-\infty ;-2)\cup (\ 2\ ;+\infty \, )\\x\in (-3\ ;\ 3\ )\end{array}\right\ \ \ \Rightarrow \ \ \ x\in (-3\ ;\, -2\ )\cup (\ 2\ ;\ 3\ )\\\\\\D(f)= (-3\ ;\, -2\ )\cup (\ 2\ ;\ 3\ )$

3 votes Thanks 1

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