When x > 3, the inequality x + 1 / (x-1) - a ≥ 0 holds, and the value range of real number a is obtained

When x > 3, the inequality x + 1 / (x-1) - a ≥ 0 holds, and the value range of real number a is obtained

When x > 3, the inequality x + 1 / (x-1) - a ≥ 0 holds, that is, when x > 3, the inequality x + 1 / (x-1) ≥ a holds, that is, when x > 3, the inequality a ≤ x + 1 / (x-1) holds. When x > 3, the inequality a ≤ (x-1) + 1 / (x-1) + 1 holds, that is, f (x) = (x-1) + 1 / (x-1) + 1 (x > 3), that is, a ≤ f (x) = (x-1

-Solving quadratic inequality of one variable with x square + 2x-2 greater than 0

-x^2+2x-2>0
x^2-2x+2

Why is the solution set of quadratic inequality of one variable all real numbers if a is greater than 0 and the discriminant of root is less than 0

A negative sum of 0 is also a real number

Solve the quadratic inequality x 2 + ax + 1 > 0 (a is a real number)

When △ = 0, the solution is a = ± 2. The inequality x2 + ax + 1 ＞ 0 is changed into (x ± 1) 2 ＞ 0, and the solution is x ≠± 1. At this time, the solution set of the inequality is: {x | x ∈ R, X ≠± 1}. When △ 0, that is, when a ＞ 2 or a ＜ - 2, from x2 + ax + 1 = 0, the solution set of the inequality is x = − a ± A2 − 42. From the inequality x2 + ax + 1 ＞ 0, the solution set of the inequality is: {x | x ＞− a + A2 − 42 or X ＜ a − A2 − 42} When △ 0, i.e. - 2 < a < 2, the solution set of inequality x2 + ax + 1 > 0 is ±. To sum up, we can see that: ① when △ = 0, the solution set of the original inequality is: {x | x ∈ R, X ≠± 1}. ② when △ 0, i.e., a > 2 or a < - 2, the solution set of the original inequality is: {x | x ＞ a + A2 − 42 or X ＜ a − A2 − 42}. ③ when △ 0, i.e. - 2 < a < 2, the solution set of inequality x2 + ax + 1 > 0 is ∅．