Question

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

Answer 1

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;∞)

Answer 2

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

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

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

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

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

$log_{0.5}(3 x^{2} +6x-x-2)\ \textless \ log_{0.5}( x^{2} -x+18x-18)$

$3 x^{2} +6x-x-2\ \textgreater \ x^{2} -x+18x-18$

$3 x^{2} +6x-x-2- x^{2} +x-18x+18\ \textgreater \ 0$

$2 x^{2} -12x+16\ \textgreater \ 0$

$x^{2} -6x+8\ \textgreater \ 0$

$D=(-6)^2-4*1*8=36-32=4$

$x_1= \frac{6+2}{2}=4$

$x_2= \frac{6-2}{2}=2$

      +                  -                   +

-------------(2)-------------(4)-------------

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С учётом ОДЗ получаем

ответ:  $(1;2)$ ∪ $(4;+$ ∞ $)$