The solution is about x inequality ax2 a 1 x 1 0

The answers are: a>1, in the range of (1 a,1);

When a=1, there is no range less than 0;

0a<0, then the range is (negative infinity, 1 a) and (1, positive infinity) The solution process: ax -(a+1)x+1 0 can be factored.

is (ax-1)(x-1)<0.

a>0, the function image.

The opening is upward, less than 0 is between 1 and 1 a, a>1, 1 a<1, then the range is (1 a, 1), when a<1, 1 a>1, then the range is (1, 1 a), when a=1, there is no range less than 0.

When a<0, the opening of the function image is downward, and a must be less than 0, then the range is (negative infinity, 1 a) and (1, positive infinity).

When a=0, we get -x+1<0, which is x>1.

When a≠0, (ax-1)(x-1) <0, and if a<0, then x<1 a, or x>1;

If 01, then 1 a1}; When a=0; When 01 is .

If a=0, the original inequality can be reduced to: x+1 0, and the solution: x -1 if a 0, the original inequality can be bridged as: (ax-1)(x+1) 0, 1 a -1, x stupid before 1 a or x -1

If -1 a 0, then 1 a -1, the solution does not get: 1 a x -1 If a -1, then 1 a -1, the solution gets: -1 x 1 a

Solution: If a=0, the original inequality can be reduced to: x+1 0, and the solution: x -1 is guessed if a 0, and the original inequality can be reduced to Solu: (ax-1)(x+1) 0, 1 a -1, x 1 Shiqing with a or x -1

If -1 a 0, then 1 a -1, solution: 1 a x -1 If a -1, then 1 a -1, solution: -1 x 1 a

If a=0, then -2x<0, x>0

If a>0

then x 2 - (2 a) x + 1<0

The opening is upward, if the discriminant formula 4 a 2-4<0,4 a 2<4,1 a 2<1 is a 2>1, from a > 0, so a > 1

At this time, x 2-(2 a)x+1 is evergreen at 0, and there is no solution.

If a=1, the discriminant is equal to 0

Then (x-1) 2<0, no solution.

If 00 opens upwards, if the discriminant formula = 4 a 2-4<0, a 2>1, from a < 0, so a <-1

At this time, x 2 - (2 a) x + 1 Evergrande is 0

If a=1, the discriminant is equal to 0

Then (x+1) 2>0, x is not equal to -1

If -11+ (1-a 2).

In summary. a<-1,x∈r

a=-1,x≠-1

11+√(1-a^2)

a=0,x>0

0

Solution: ax -(2a+1)x+2<0

Equivalent to (ax-1)(x-2)<0

When a<0, the original inequality high car iso-burn oak price is (x-1 a)(x-2)>0a<0, 1 a<0<2, and the solution set of the inequality is;

When a=0, the solution set of inequalities is;

When 02, the solution set of inequalities is;

When a=0, the solution set of inequalities is;

When 01 2, the solution set of inequalities is {x|1/a

Commenting on Qi Duan's remarks.

The key is to discuss the size of the two sticks.

When a=0, the original formula is: x-2>0, solution set.

When a≠0, (ax-1)(x-2)=0, the two roots are 1 a, and 2 when 1 a>2, that is, 01 2, there are two cases:

a<0, the original is (x-1 a)(x-2)>0, the solution set is a>1 2, the original is (x-1 a)(x-2)<0, and the solution set is.

When a=0, the solution set of inequality is (2,+infinity), when a is not =0, let (ax-1)(x-2)=0, and x does not =2 solve x=1 a, when a=1 2, the solution of inequality calls Qi an empty set; When a>1 2, the solution set of inequalities is 1 a,

When a 0, there is (x 2) 0, which is x 2

If a0, when 0 is good for a, then 2 x 1 a, when a, x is an empty set, when a, 1 macro sock or a x 2

I hope my shield can help you.

There are two situations for this to be discussed, when a>0, x>1 (a+1 a) is solved, and when a<0, the repentance is a<1 (a+1 a).