Question

(0,8)^2x-x^2≥1 помогите пожалуйста

Answer 1

$(0.8)^{2x-x^2}geq 1 \\ (0.8)^{2x-x^2}geq (0.8)^0 \\ 2x-x^2leq 0 \\ x(2-x)leq0 \\ left { {{x leq 0}} atop {x geq 2} right. \\ xin (-infty;0] cup [2;+infty)$