If the solution set of inequality MX2 + MX-1 > 0 about X is an empty set, the value range of real number m is obtained

If the solution set of inequality MX2 + MX-1 > 0 about X is an empty set, the value range of real number m is obtained

(1) When m = 0, the original inequality is: - 1 > 0, which is not true, and the solution set of X is empty;
(2) When m ≠ 0, the original inequality is a quadratic inequality of one variable. If the solution set is an empty set, then the discriminant
△=m＾2+4m

Given the set a = {X - 2 ≤ x ≤ 5}, set B = {x m + 1 ≤ x ≤ 2m-1}, and a ∩ B = a, try to find the value range of real number M

Because a ∩ B = a, then set a is a subset of set B. therefore, for set B, we must:
m＋1≤2m－1
m≥2
In addition, if set a is a subset of set B, then:
(1)m＋1≤－2
(2)2m－1≥5
The solution is that m does not exist
Thus, the value of m does not exist and has no solution

If the inequality f (x) > A-X holds, the value range of real number a can be obtained

∵ f (x) = | x | + | x-3 |, ∵ f (x) = − 2x + 3, X ＜ 03, 0 ≤ x ≤ 32x − 3, X ＞ 3 (1) x ＜ 0, - 2x + 3 ＞ A-X, ∵ a ＜ - x + 3, when inequality f (x) ＞ A-X is constant, when a ＜ 3 (2) 0 ≤ x ≤ 3, when 3 ＞ A-X, ∵ a ＜ x + 3, ∵ when inequality f (x) ＞ A-X is constant, a ＜ 3 (....)

If one part of a + one part of B = 4, then 2A + 2b-7ab part of a-3ab + B =?

1 / A + 1 / b = 4, (a + b) / (AB) = 4, a + B = 4AB, so (a-3ab + b) / (2a + 2b-7ab) = (AB) / (AB) = 1