If the solution set of inequality | x2-8x + a | ≤ x-4 is [4,5], then the value of real number a is equal to______ ．

If the solution set of inequality | x2-8x + a | ≤ x-4 is [4,5], then the value of real number a is equal to______ ．

The solution set of the inequality | x2-8x + a | ≤ x-4 is [4,5], so 4,5 is the root of the corresponding equation of the inequality, | 16 − 32 + a | ≤ 4 − 4 | 25 − 40 + a | ≤ 5 − 4, the solution is a = 16, so the answer is: 16

Let P: the solution set of inequality x + | x-2m | > = 2 be all real numbers; proposition q: the solution set of inequality x2-x + 1 - (1 / 4) m > 0 is also all real numbers;
If there is only one correct proposition p and Q, try to find the range of real number M

For P: inequality x + | x-2m | ≥ 2
If x-2m ≥ 0, then x + | x-2m | = x + x-2m = 2 (x-m) ≥ 2, X ≥ m, then the solution is x ≥ Max {m, 2m}
If x-2m (M-3) / 4
Because (x-1 / 2) ^ 2 ≥ 0, when (M-3) / 4