Given that the solution set of inequality (2a-b) x + a-5b 〉 0 about X is x 〈 7 / 10, find the solution set of inequality ax 〉 B about X

Given that the solution set of inequality (2a-b) x + a-5b 〉 0 about X is x 〈 7 / 10, find the solution set of inequality ax 〉 B about X

(2a-b)x+a-5b〉0
(2a-b)x>5b-a
x

Given that the solution set of inequality (2a-b) x + a-5b > 0 about X is x < 10 / 7, find the solution set of inequality ax > B

Because the solution of (2a-b) x + a-5b > 0 is x0
bx>(10/7)b
Because X0
So x

Given that the solution of inequality (2a-b) x + a-5b > 0 about X is x ＜ & nbsp; 107, then the solution of AX + b > 0 is ()
A. x＞ −35B. x＜− 35C. x＞ 35D. .x＜ 35

The solution of (2a-b) x ＞ - A + 5b, ∵ inequality (2a-b) x + a-5b ＞ 0 is x ＜ & nbsp; 107, ∵ 2a-b ＜ 0, and − a + 5b2a − B = 107. By sorting out the equation, B = 35A, substituting 2a-b ＜ 0, 2a-35a ＜ 0, the solution set ax + 35A ＞ 0 of a ＜ 0, ∵ ax + B ＞ 0, that is, X ＜ - 35

The solution set of inequality ax-x > b is X

ax-x>b
(a-1)x

If the solution of the inequality ax-a ＞ X-1 is x ＜ 1, then the value range of A
emergency

ax-a＞x-1
ax-x>a-1
x(a-1)>a-1
Divide both sides by A-1
The solution is X

If the solution set of inequality AX-1 > x-a about X is X

The solution is AX-1 > x-a
The results show that ax-x > 1-A
That is, (A-1) x > 1-A
From the solution set of inequality AX-1 > x-a to x1-a
x＜（1-a）/（a-1）=-（a-1）/(a-1）=-1
That is, X ＜ - 1
That is to say, the problem can satisfy A-1 < 0
That is a < 1

If the solution set of the inequality ax square - ax + 3 > 0 about X is r, then the value range of a is r
Urgent need
If the solution set of the inequality ax square - 2aX + 3 > 0 about X is r, then the value range of a is 2aX. Sorry to type wrong

Analysis
△=b²-4ac

It is known that the solution set of the inequality ax ^ 2 + ax + 1 > 0 about X is r, and the value range of a is obtained

① When a ＜ 0, f (x) = ax & # 178; + ax + 1, the opening of the image is downward, then there must be x, so that f (x) ＜ 0, so it is not satisfied. ② when a = 0, f (x) = 1 ＞ 0, it is constant. ③ when a ＞ 0, f (x) = ax & # 178; + ax + 1, the opening of the image is upward. If f (x) ＞ 0, it is constant, that is, there is no intersection point with X axis. Then △ = A & # 178; -

Known 0

F (x) = x ^ 2 + ax + 3-A > 0
The axis of symmetry is
x=-a/2
When - A / 20
Minimum value = f (0) = 3-A > 0
∴0

If there are only three positive integer solutions to the inequality 3x-a ≤ 0, then the value range of a is larger______ ．

Then the positive integer solution is: 1, 2, 3. So 3 ≤ A3 < 4. The solution is: 9 ≤ a < 12. So the answer is: 9 ≤ a < 12