помогите с В7,пожалуйста. Created by dariarakovich2004. algebra-ru

Объяснение:ОДЗ: x+7≠0 x≠-7Пусть -∞__-__-1/2__+__0__-__1__+__+∞t∈(-1/2;0)U(1;+∞).t>0 t≠1 ⇒t∈(1;+∞) ⇒Ответ: x=-8.

помогите с В7,пожалуйста...

dariarakovich2004 @dariarakovich2004

August 2022 1 0 Report

помогите с В7,пожалуйста

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sangers1959

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Объяснение:

$(\frac{1}{2} )^{\frac{2}{x+7}}+(\frac{1}{4} )^{\frac{2}{x+7}}\leq 2*(\frac{1}{8})^{\frac{2}{x+7}} \\ (\frac{1}{2} )^{\frac{2}{x+7}}+(\frac{1}{2} )^{\frac{2*2}{x+7}}\leq 2*(\frac{1}{2})^{\frac{3*2}{x+7}.$

ОДЗ: x+7≠0 x≠-7

Пусть $(\frac{1}{2})^{\frac{2}{x+7}} =t>0 \ \ \ \ \Rightarrow$

$t+t^2\leq 2*t^3\\2t^3-t^2-t\geq 0\\t*(2t^2-t-1)\geq 0\\t*(2t^2-2t+t-1)\geq 0\\t*(2t*(t-1)+(t-1))\geq 0\\t*(t-1)(2t+1)\geq 0\\t_1=0\notin.\\t-1=0\\t_2=(\frac{1}{2})^{\frac{2}{x+7}}=1\ \ \ \ \Rightarrow\\(\frac{1}{2})^ \frac{2}{x+7}=(\frac{1}{2})^0\\\frac{2}{x+7} =0\\x\in \varnothing.\\2t+1=0\\2t=-1\ |:2\\t=(\frac{1}{2})^{\frac{2}{x+7}}=-\frac{1}{2}\ \ \ \ \Rightarrow\\t\notin. \\ t*(t-1)*(2t+1)\geq 0.$

-∞__-__-1/2__+__0__-__1__+__+∞

t∈(-1/2;0)U(1;+∞).

t>0 t≠1 ⇒

t∈(1;+∞) ⇒

$(\frac{1}{2})^{\frac{2}{x+7} }\geq 1\\(\frac{1}{2})^{\frac{2}{x+7}}\geq (\frac{1}{2})^0\\\frac{2}{x+7}\leq 0\\x+7$

Ответ: x=-8.

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