Решите неравенства:

А) log1/2 (2x+1) >-2

Б) (1/4) ^х'2-6< (1/2) ^10 х

А) log1/2 (2x+1) >-2

(1/2) ^ (-2) = 2x + 1

2x = 3

x = 3/2

x = 1,5

Б) (1/4) ^х'2-6< (1/2) ^10 х

(1) 2) ^ (2x^2 - 12) < (1/2) ^ (10x) Так как o< 1/2 < 1, то

2x^2 - 12 > 10x

2x^2 - 10x - 12 > 0

x^2 - 5x - 6 > 0

x1 = - 1

x2 = 6

x ∈ (-≈; - 1) (6; + ≈)