2.
log₂log₁/₂((x+1)/(x-3))≥1
ОДЗ: (x+1)/(x-3)>0 -∞___+___-1___-___3___+___+∞ ⇒ x∈(-∞;-1)U(3;+∞).
log₁/₂((x+1)/(x-3))>0 -log₂((x+1)/(x-3))>0 |÷(-1) log₂((x+1)/(x-3))<0 <br>(x+1)/(x-3)<1 (x+1)/(x-3)-1<0 (x+1-(x-3)/(x-3)<0 (x+1-x+3)/(x-3)<0 4/(x-3)<0<br> x-3<0 x<3 ⇒ <strong>x∈(-∞;-1).
log₁/₂((x+1)/(x-3))≥2
2*log₂((x+1)/(x-3))≥2 |÷2
log₂((x+1)/(x-3))≥1
(x+1)/(x-3)≥2
(x+1)/(x-3)-2≥0
(x+1-2*(x-3))/x-3)≥0
(x+1-2x+6)/(x-3)≥0
(7-x)/(x-3)≥0
-∞____-____3____+____7____-____+∞ ⇒
x∈(3;7] ОДЗ: x∈(-∞;-1) ⇒
Ответ: неравенство решения не имеет.