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