If the square of inequality X - 6x > = n is (O, 1] constant for any x, then the value range of n is?

If the square of inequality X - 6x > = n is (O, 1] constant for any x, then the value range of n is?

The axis of symmetry of the function f (x) = x ^ 2-6x is x = 3, monotonically decreasing on (- ∝, 3] and monotonically increasing on [3, + ∞). Therefore, the minimum value of F (x) on (O, 1] is f (1) = 1-6 = - 5. Because the square of the inequality x-6x > = n is constant for any x, so only the minimum value of F (x) on (O, 1] ≥ n

For any a ∈ [- 4,5], the inequality x ^ - 6x ＜ a (X-2) holds and the range of value of X is obtained

The inequality is reduced to:
(x-2)a-x^2+6x>0
Let f (a) = (X-2) A-X ^ 2 + 6x
If x = 2, then f (a) = 8 > 0 holds
When x0, the solution is: 10, the solution is: 2

Factorization: xn power - xn power - 2 power

Xn power - xn power - 2 power
=x^(n-2)(x^2-1)
=(x-1)(x+1)x^(n-2)