Помогите пожалуйста подробно решить неравенства. даю 22 балла

1
ОДЗ
x>0,x≠1,x+1>0⇒x>-1
x∈(0;1) U (1;∞)
2log^-1(x)36=log(6)x
log(6)x+log(6)(x+1)≤1
log(6)(x²+x)≤1
x²+x≤6
x²+x-6≤0
x1+x2=-1 U x1*x2=-6⇒x1=-3 U x2=2
-3≤x≤2 U x∈(0;1) U (1;∞)
Ответ x∈(0;1) U (1;2]
2
ОДЗ
x-1>0⇒x>1
2x-4>0⇒x>2
x∈(2;∞)
log(2)[(x-1)²/(2x-4)]>1
(x-1)²/(2x-4)>2
(x-1)²/(2x-4)-2>0
(x²-2x+1-4x+8)/(2x-4)>0
(x²-6x+9)/(2x-4)>0
(x-3)²/(2x-4)>0
x-3=0⇒x=3
2x-4=0⇒x=2
_ + +
--------------(2)--------------(3)--------------------
Ответ x∈(2;3) U (3;∞)
3
ОДЗ
x-1>0⇒x>1
x+1>0⇒x>-1
(x+1)/(x-1)>0⇒x<-1 U x>1
(x+1)/(x-1)≠1⇒x+1≠x-1
x∈(1;∞)
log[(x+1)/(x-1)]2>log(2)(x+1)-log(2)(x-1)
log[(x+1)/(x-1)]2>log(2)[(x+1)/(x-1)]
log[(x+1)/(x-1)]2>1/log[(x+1)/(x-1)]
log[(x+1)/(x-1)]2=a
a-1/a>0
(a²-1)/a>0
(a-1)(a+1)/a>0
a=1 a=-1 a=0
_ + _ +
---------------(-1)-----------(0)--------------(1)------------
-1(x-1)/(x+1)<2 (x-1-2x-2)/(x+1)<0 (x+3)/(x+1)<0⇒-3<x<-1 не удов усл a>1⇒log[(x+1)/(x-1)]2>1
(x+1)/(x-1)<2 (x+1-2x+2)/(x-1)<0 (3-x)/(x-1)<0 x<1 U x>3 U x>1
Ответ x∈(3;∞)