(log по 5 (25х^2) + 48) / (log по 5 от х^2 - 49) >= -1. Created by Cложно. algebra-ru

ОДЗ: x ≠ 0 (ОДЗ логарифма)Пусть Учитывая ОДЗ, получаем ответ:

(log по 5 от х^2 - 49) >= -1...

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ОДЗ: x ≠ 0 (ОДЗ логарифма)
$\log_{5}{x^2}-49 \neq 0 \\ \log_{5}{x^2} \neq 49 \\ x^2 \neq 5^{49} \\ x \neq \pm5^{24}\sqrt{5}$

$\frac{\log_{5}{(25x^2)}+48}{\log_{5}{x^2}-49} \geq -1 \\ \frac{2\log_{5}{(|5x|)}+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2(\log_{5}{5}+\log_{5}{|x|})+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2(1+\log_{5}{|x|})+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2+2\log_{5}{|x|}+48}{2\log_{5}{|x|}-49} \geq -1 \\ \frac{2\log_{5}{|x|}+50}{2\log_{5}{|x|}-49} \geq -1$
Пусть $2\log_{5}{|x|}=t$
$\frac{t+50}{t-49} \geq -1 \\ \frac{t+50}{t-49} + 1 \geq 0 \\ \frac{2t+1}{t-49} \geq 0 \\ \left [ {{t \leq -0.5} \atop {t\ \textgreater \ 49}} \right. \\ \left [ {{2\log_{5}{|x| \leq -0.5}} \atop {2\log_{5}{|x|}\ \textgreater \ 49}} \right. \\ \left [ {{\log_{5}{|x|} \leq -0.25} \atop {\log_{5}{|x|}\ \textgreater \ \frac{49}{2} }} \right. \\ \left [ {{\log_{5}{|x|} \leq \log_{5}{5^{- \frac{1}{4} }}} \atop {\log_{5}{|x|}\ \textgreater \ \log_{5}{5^{\frac{49}{2}}} }} \right.$
$\left [ {{|x| \leq \frac{1}{ \sqrt[4]{5} } } \atop {|x|\ \textgreater \ \sqrt{5^{49}} }} \right. \\ \left [ { -\frac{ \sqrt[4]{125} }{5}\leq {x \leq \frac{ \sqrt[4]{125} }{5} } \atop { \left [ {{x\ \textless \ -5^{24} \sqrt{5} } \atop {x\ \textgreater \ 5^{24} \sqrt{5} }} \right. }} \right.$

Учитывая ОДЗ, получаем ответ:
$x\in(-\infty; -5^{24} \sqrt{5})\cup[-\frac{ \sqrt[4]{125} }{5}; 0)\cup(0; \frac{ \sqrt[4]{125} }{5}]\cup(5^{24}\sqrt{5}; +\infty)$

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Cложно Спасибо большое)

sedinalana Log(5)(25x²)+48=log(5)25+log(5)x²+48=2+2log(5)x+48=log(5)x+50
log(5)x²-49=2log(5)x-49
----------------------------------------------
x>0
(2log(5)x+50)/(2log(5)x-49)≥-1
(2log(5)x+50)/(2log(5)x-49)+1≥0
(2log(5)x+50+2log(5)x-49)/(2log(5)x-49)≥0
(4log(5)x+1)/(2log(5)x-49)≥0
log(5)x=t
(4t+1)/(2t-49)≥0
t=-1/4 t=49/2
+ _ +
-----------[-1/4]----------------(49/2)-------------
t≤-1/4⇒log(5)x≤-1/4⇒ $x \leq 1/ \sqrt[4]{5}$
t>49/2⇒log(5)x>49/2⇒x> $5 ^{24} \sqrt{5}$
Ответ ∈(0; $1/ \sqrt[4]{5}$ ] U ( $5 ^{24} \sqrt{5}$ ;∞)

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