Решите неравенство: log0,5 (3x-1)-log0,5 (x-1)

August 2021 1 11 Report

Решите неравенство: log0,5 (3x-1)-log0,5 (x-1)<log0,5 (x+18)-log0,5 (x+2)

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Verified answer

ОДЗ
{3x-1>0⇒x>1/3
{x-1>0⇒x>1
{x+18>0⇒x>-18
{x+2>0⇒x>-2
x∈(1;∞)
log(0,5)[(3x-1)/(x-1)]<log(0,5)[(x+18)/(x+2)]
(3x-1)/(x-1)>(x+18)/(x+2)
(3x-1)/(x-1)-(x+18)/(x+2)>0
(3x²+6x-x-2-x²-18x+x+18)/[(x-1)(x+2)]>0
(2x²-12x+16)/[(x-1)(x+2)]>0
2(x²-6x+8)=0
x1+x2=6 U x1*x2=8⇒x1=2 U x2=4
x-1=0⇒x=1
x+2=0⇒x=-2
+ _ + _ +
---------------(-1)-------------(1)--------------(2)----------(4)------------
//////////////////////////////////////////////////////////
x∈(1;2) U (4;∞)

mukus13 mukus13

$log_{0.5}(3x-1)-log_{0.5}(x-1)\ \textless \
log_{0.5}(x+18)-log_{0.5}(x+2)$

ОДЗ:
$\left \{ {{3x-1\ \textgreater \ 0} \atop
{x-1\ \textgreater \ 0}} \atop {x+18\ \textgreater \ 0\atop{x+2\ \textgreater \
0} \right.$

$\left \{ {{x\ \textgreater \ \frac{1}{3} }
\atop {x\ \textgreater \ 1}} \atop {x\ \textgreater \ -18\atop{x\ \textgreater
\ -2} \right.$
$x$ ∈ $(1;+$ ∞ $)$

$log_{0.5}(3x-1)+log_{0.5}(x+2)\ \textless \
log_{0.5}(x+18)+log_{0.5}(x-1)$