(1\5)^х-1 + (1\5)^x+1<= 26
(1/5)ˣ⁻¹+(1/5)ˣ⁺¹≤26 5*(1/5)ˣ+(1/5)ˣ/5≤26 |×5 25*(1/5)ˣ+(1/5)ˣ≤26*5 26*(1/5)ˣ≤26*5 |÷26 (1/5)ˣ≤5 (1/5)ˣ≤(1/5)⁻¹ x≥-1 Ответ: x∈[-1;+∞).