Question

log0.5(x^2-1)<=log0.5(2x+7)

Answer 1

0,5<1⇒x²-1≥2x+7        ОДЗ    х²-1>0    ⇒|x|>1

x²-2x-8≥0                                 2x+7>0 ⇒x>-3,5   x 

y=x²-2x-8

x²-2x-8=0

D1=1+8=9

x1=1+3=4     x2=1-3=-2

(x+2)(x-4)≥0

_____+____-2____-____4____+_____

x∈(-3,5 -2]∨[4 ∞)