The inequality about x x 1 ax has only three integer solutions, then the range of values of the real

a-1)x²+2x-1＜0

1. a=1, which is y=(a-1)x +2x-1 is a one-time equation with an infinite number of integer solutions.

2. a-1 0, that is, a 1, y is a quadratic function. Intersection with y-axis: -1, axis of symmetry: -1 (a-1)>0, opening downward.

The approximate image is: (I don't know if there is an intersection with the x-axis, but it doesn't matter).

So there are countless integer solutions.

3. a-1>0, i.e., a>1, y is a quadratic function. Intersection with y-axis: -1, axis of symmetry: -1 (a-1)<0, opening up.

The approximate image is:

x=0 is an integer solution, and there can only be two integer solutions, and the other two solutions must be -2 and -1 as can be seen from the graph. (So the intersection of y and x-axis is between 0,1 and the other is between -2,-3).

When x=0, y<0Solution: A is R

When x=1, y>0 is solved: a>0

When x=-3, y>0 solves: a>16 9

When x=-2, y<0 solves: a<9 4

In summary: 16 9

First of all, when a 0, it is clear that x has an infinite number of integer solutions.

When a 0, let t = root number a, a = t, t 0 [because the root number is not easy to type, it is easy to type and easy to see].

When x=0, the inequality holds.

When x≠0 (x-1) x t

1-(1x)t or 1-(1x)-t

1 x) 1-t or (1 x) 1+t

When 1-t 0, x 0 has an infinite number of integer solutions, so 1-t 0, i.e. t 1 gives 1 (1-t) x 1 (1+t), obviously 1 (1+t) 1, so the integer solution is 0, -1, -2

So 1 (1-t) -2 gives t (1,3 2), a (1,9 4).

You don't need to change the yuan when you write, just use the root number A instead of t].

Solution: Inequalities can be simplified as.

a-1)x²+2x-1<0

1.If a=1, the inequality can be reduced to x<1 2, and there are an infinite number of integer solutions that do not match the meaning of the question, so a≠1.

2.If a>1, the inequality can be reduced to [x+1 ( a-1)][x-1 ( a+1)]<0, and 1 (1- a)ax has and only three integer solutions, therefore.

2 1 (1+ a)-1 (1- a)<3,2 2 a (a-1)<3,3- 5) 2 a<(11-2 10) 9,Disagreement with a>1, discarded.

11+2 10) 90,1) If a<0, then the inequality is constant, which does not match the meaning of the title.

2) If 00, x<1 (1+ a) or x>1 (1- a), it does not match the title.

In summary, the value range of a is (11+2 10) 9

x=1 is solved by the three positive integer bucket numbers in the problem; x=2;x=3.So 3

If the inequality about x, x a, there are only three positive integers, and the hungry spine stall is a mess.

Then x can takeWhereas. x≤a

Then there are 3 a

For the inequality of x, 0 x+1a+3, find the range of values of a.

According to 0 x+1<3, we can know that the value range of x is 0 xa+3, and we can deduce a < 2x - 3, in other words, the value range of a is a 2x - 3. To sum up, the range of which finch wisdom value of a is a 1.

From the meaning of the title, it can be seen that x is less than or equal to -a and there are only two positive integer solutions, so -a can only be greater than or equal to 2 and less than 3, so a is less than or equal to -2 and greater than -3

Known: x-a 0, x is greater than 0 and there are only 2 positive integer solutions.

Seek: xx-a 0

When x is next to aa=1, x has 1 solution; When a=2, x has 2 solutions; A=3, x has Yu Qiye shouting 3 solutions. 2≤a

The inequality is deformed as: x a 3

Since there are only two positive integer solutions to the inequality, 2 a 3 3, so that 6 a 9 If 3 is in the solution of the positive integer, then there must be a positive integer solution in the solution of the inequality, so that there are 3 solutions to the positive integer instead of two. If there are more than 4 numbers in the positive integer solution, then there are more positive integer solutions in the solution, and none of them meet the condition that the problem has only two positive integer solutions.

Another way is to draw the solution set of the inequality on the number line, first select the point a 3 on the number line, take this point as a solid point, draw the solution set of the inequality to the left, and make the solution set contain only two positive integer solutions, it is not difficult to find that these two positive integers can only be 1 and 2, and at the same time, it can be found that a 3 can only fall in the area of 2 to 3, and can include 2 but not 3, that is, the origin of 2 a 3 3 above.

x<=a 3 and only two positive integer solutions must be 1 2 so 2 2 2 so <2 = a 3<3

6<=a<9

x<2, i.e., 3 integer solutions do not take 2, but start with 1 less than 2.

So 3 integers are 1, 0, -1

And because x>a, that is, you can't get a, but a can't be greater than -2, and if it's greater than it, you can take 4.

So -2=

5-2x>1=>x<2

x-a≥0=>x≥a

Since there are only four integer solutions in the inequality group, the value of x can only be 1, 0, -1, -2, so the range of a is (-3, -2).