Решение:
36 – 12х + х2 < √10(x – 6)
х2 – 12х + 36 < √10(x – 6)
х2 – 2·6·х + 62 < √10(x – 6)
(x – 6)2 < √10(x – 6)
(x – 6)2 – √10(x – 6) < 0
(x – 6)·(x – 6 – √10) < 0
метод интервалов:
(x – 6)·(x – 6 – √10) = 0
x – 6 = 0
x = 6
или
x – 6 – √10 = 0
x = 6 + √10
х ∈ (6; 6 + √10)
Ответ: (6; 6 + √10)