Решение:
81 – 18х + х2 < √2(x – 9)
х2 – 18х + 81 < √2(x – 9)
х2 – 2·9·х + 92 < √2(x – 9)
(x – 9)2 < √2(x – 9)
(x – 9)2 – √2(x – 9) < 0
(x – 9)·(x – 9 – √2) < 0
метод интервалов:
(x – 9)·(x – 9 – √2) = 0
x – 9 = 0
x = 9
или
x – 9 – √2 = 0
x = 9 + √2
х ∈ (9; 9 + √2)
Ответ: (9; 9 + √2)