(5²)¹⁾⁵⁺ˣ ≤ 80*5ˣ⁺¹ - 275
5³ * 5ˣ ≤ 80*5ˣ⁺¹ - 275 |: 5
5²*5ˣ ≤ 16*5ˣ⁺¹ - 55
5ˣ = t
25t² -80t +55 ≤ 0
5t² -`16t +11 ≤ 0
t₁ = 1 5ˣ = 1, ⇒ х = 0
t₂ = 11/5 5ˣ = 11/3, ⇒ хlg5 = lg11 - lg5, ⇒ x = (lg11 - lg5)/lg5
-∞ 0 (lg11 - lg5)/lg5 +∞
+ - +
IIIIIIIIIIIIIIIIIIIIIIIIIII]
Ответ: х∈[0; (lg11 - lg5)/lg5]