Question

((1)/(7))^(2x-1)-8((1)/(7))^(x)+1<=0

Answer 1

(1/7)⁽²ˣ⁻¹⁾-8*(1/7)ˣ+1=0

(1/7)⁻¹*(1/7)²ˣ-8*(1/7)ˣ+1=0

7*(1/7)²ˣ-8*(1/7)ˣ+1=0

Пусть (1/7)ˣ=t>0     ⇒

7t²-8t+1=0      D=36      √d=6   ⇒

t₁=(1/7)ˣ=1          x₁=0

t₂=(1/7)ˣ=1/7      x₂=1.

Ответ: x₁=0        x₂=1.