1
ОДЗ
{(1/2-x)(x-4)≥0
{(2x+0,2)/(x+1)>0
x=1/2 x=5 x=-0,1 x=-1
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x∈[1/2;4]
√(1/2-x)(x-4)≥0⇒log(2)[(2x+0,2)/(x+1)]≤0
(2x+0.2)/(x+1)≤1
(2x+0,2-x-1)/(x+1)≤0
(x-0,8)(x+1)≤0
x=0,8 x=-1
-1Ответ x∈[0,5;0,8]
2
ОДЗ x≠1
1){(x-1/2)(x-3)>0⇒x<0,5 U x>3
{log(2)|x-1|>0⇒|x-1|>1⇒x-1<-1 U x-1>1⇒x<0 U x>2
x∈(-∞;0) U (3;∞)
2){(x-1/2)(x-3)<0⇒0,5 <x<3<br>{log(2)|x-1|<0⇒|x-1|<1⇒-1<x-1<1⇒0<x<2<br>x∈(0;1) U (1;2)
Ответ x∈(-∞;0) U (0;1) U (1;2) U (3;∞)