Given proposition p: in X ∈ [1,2], inequality x2 + AX-2 ＞ 0 holds; proposition q: function f (x) = log13 (x2 − 2aX + 3a) is a decreasing function on interval [1, + ∞). If proposition "P ∨ Q" is a true proposition, the value range of real number a is obtained

Let g (x) = 2x − x, then G (x) is a decreasing function on [1,2], and G (x) max = g (1) = 1, then a > 1 if the proposition p is true; Let f (x) = log13 (x2 − 2aX + 3a) be a decreasing function on the interval [1, + ∞), and U (x) = x2 − 2aX + 3A be an increasing function on [1, + ∞) The function U (x) = x2 − 2aX + 3a ＞ 0 is constant on [1, + ∞]. That is, if proposition q is true, then - 1 ＜ a ≤ 1. If proposition "P ∨ Q" is true, then there is p true Q false or P false Q true or P, q are true propositions. If P true Q false, then there is a > 1. If P false Q true, then there is - 1 ＜ a ≤ 1. If P, q are true propositions, then there is no A. In conclusion, the value range of real number a is a ＞ - 1 ．

Known proposition: P: "any x ∈ (0, + ∞), inequality ax ≤ x ^ 2-A is always tenable", proposition q: "1 is an inequality about X"
(x-a) (x-a-1) ≤ 0, if there is only one true proposition in two propositions, what is the value range of real number a?

When the proposition: P: "any x ∈ (0, + ∞), the inequality ax ≤ x ^ 2-A holds", the range of solution a is a

We know the following two propositions: P: for all x? R, the inequality ax > = 2 √ X - 1 holds; Q: 1 is the inequality (x-a) (x-a-1) = 2 √ X - 1 holds; Q: 1 is the inequality (x-a) (x-a-1) about X

P :a>=1
Q: A > = 1 or a

Solve the system of inequalities 3 (x + 1) < 2x + 4, (x-1) / 3 > = x / 2-1 and express its solution set on the number axis

If the solution set of a (x-1) > x + 1-2a is x < - 1, then the value range of a is______ ．

The solution set of (A-1) x ＞ 1-A, ∵ a (x-1) ＞ x + 1-2a is x ＜ - 1, ∵ A-1 ＜ 0, i.e. a ＜ 1

The solution set of inequality x + | x-2a | > 1 is the value range of R for a
The solution set of inequality x + ｜ x-2a ｜ > 1 is the value range of R for a
Give the process

The solution set of X + ︱ x-2a ︱ 1 is r,
x> When x = 2A, x + x-2a-1 = x + x-2a-1 = 2x-2a-1 > 2 * 2a-2a-1 = 2a-1 > 0
=>a>1/2
x0
=>a>1/2
To sum up, a > 1 / 2

If x ＜ a + 1 x ＞ 2a-1, then the value range of a is
If inequality system
x＜a+1
x＞2a-1
If there is a solution, then the value range of a is

2a-1=2

If the solution set of inequality (2a-1) x < 2 (2a-1) is x > 2, then the value range of a is ()
A. a＜0B. a＜12C. a＜-12D. a＞-12

The solution set of ∵ inequality (2a-1) x ∵ 2 (2a-1) is x ∵ 2, ∵ inequality sign change, ∵ 2a-1 ∵ 0, ∵ a ∵ 12

Known inequality x-2a greater than or equal to 1 solution set is x greater than or equal to - 5, find a value range, fast ah

X-2a > = 1 leads to x > = 2A + 1
And because x > = - 5
According to the same big, we get - 5 > = 2A + 1
So we get a

If the solution set of X − 2A ＞ 02 (x + 1) ＞ 14 − x is x ＞ 2a, then the value range of a is ()
A. a＞4B. a＞2C. a=2D. a≥2

X − 2A ＞ 0, ① 2 (x + 1) ＞ 14 − x, ② the solution set of the inequality system X − 2A ＞ 02 (x + 1) ＞ 14 − x is x ＞ 2a, ｜ 2A ≥ 4, a ≥ 2, that is, the value range of a is a ≥ 2, so D is selected

Given proposition p: in X ∈ [1,2], inequality x2 + AX-2 ＞ 0 holds; proposition q: function f (x) = log13 (x2 − 2aX + 3a) is a decreasing function on interval [1, + ∞). If proposition "P ∨ Q" is a true proposition, the value range of real number a is obtained