If the system of inequalities x − 1 ＞ a2x − 4 ＜ 2A about X has a solution, then the value range of real number a______ ．

If the system of inequalities x − 1 ＞ a2x − 4 ＜ 2A about X has a solution, then the value range of real number a______ ．

X − 1 ＞ a2x − 4 ＜ 2A ⇔ A2 ＜ x ＜ 2a, so the original inequality has a solution, that is, A2 ＜ 2a, the solution is 0 ＜ a ＜ 2, so the answer is: (0, 2)

The solution set of inequality a (x-1) > x + 1-2a is x ＞ - 1, and the value range of a is
It's X-1, not X-1

a(x-1)＞x+1-2a
The solution set of (A-1) x > 1-A is x > 1
a-1>0
a>1

Known inequality system x > A + 2x

That is a + 2

If the inequality 2x ^ 2 + ax + 1 ≥ 0 holds on the interval x ∈ (0, + ∞), then the value range of a is
If the opening of the function is upward, it only needs △≤ 0,
△=a2 - 4×2×1=a2 - 8 ≤0
-2√2 ≤ a ≤ 2√2
Why is this wrong? Why should we discuss the case of △ 0? Please analyze,

If we want to discuss △ 0, we only need x ∈ (0, + ∞)
When △≤ 0, △ = a2 - 4 × 2 × 1 = a2 - 8 ≤ 0
-2√2 ≤ a ≤ 2√2
If △ 0, a ＞ 2v2, or a ＜ - 2v2, because f (0) = 1 ＞ 0, only the axis of symmetry - A / 4 ＜ 0 is needed
a＞0
So a > 2v2
As shown in the summary
a＞＝－2v2

If the solution sets of inequality x ^ 2 + ax-a-2 > 0 and 2x ^ 2 + 2 (2a + 1) x + 4A ^ 2 + 1 > 0 are a and B respectively
Find the value range of the real constant a such that a = R and B = R have at least one set

A=R =>a^2-4(-a-2)a^2+4a+8=(a+2)^2+4[2(2a+1)]^2-4*2*(4a^2+1)=16a^2+16a+4-32a^2-8=-16a^2+16a-4=-4(2a-1)^2

ax²-(4a+1)x+3a+1

Ax & # 178; - (4a + 1) x + 3A + 1 = (ax-3a-1) (x-1) 3A + 1, i.e. a (3a + 1) / A or X

If the solution set of A0 is

Inequality (x-4a) / (x + 5a) > 0
Equivalent to (x-4a) (x + 5a) > 0
∵a4a
∴x-5a
The solution set of the original inequality is (- ∞, 4a) U (- 5A, + ∞)

Inequality a & # 178; + B & # 178; + 2 ≥ 2A + 2B why?

Finding the solution set of inequality x * 2-4a + 3A * 2 < 0 about X

x^2-4a+3a^2＜0
x^20
0

A and B at the same table are arguing about a problem. A says: "5A > 4a." B says no. please tell me who gives an example of why

When a = 0, 5A = 4A
When A4a
Give me points!