Because of ax? The solution set of +bx+ 1 > 0 is {x |- 1

So {a

-b/a=- 1 + 1/3 =-2/3

1/a = - 1/3

The solution is {a=-3.

b= 2

So ab=-6

The first is a<, a; , one

2.

Solution: y= 1-6x-x? =-(x？ +6x- 1)=-[(x+3)？ - 10]= 10-(x+3)？ ,

Because (x+3)? ≥0,

So 10-(x+3)? ≤ 10，M = { y | y≤ 10 }；

Similarly, y=5+2x-x? =6-(x- 1)？

Because (x+ 1)? ≥0,

So 6-(x+ 1)? ≤ 10，M = { y | y≤6 }；

To sum up, we can see that n is indeed included in m.

3.

That is to say, 3x+ 1 = x 2-x+a+ 1, x 2-4x+a = 0 has a solution, that is, 4 2-4a > = 0a ≤ 4. Since A is a positive integer, A takes the value of 1, 2,3,4.

Description: A∩B≠ empty set means that A and B have the same elements.

4.

Let f(x)=kx? +( 1-k？ )x-kδ=( 1-k？ )? +4k？ =( 1+k？ )? >0

1. If k=0, it is not valid.

2. If k < 0, the parabola opens downward, it can be found that only F (-2) > 0 and F (2) > 0 are needed to satisfy the problem conditions.

Solution: k does not exist.

3. If k > 0, the opening is upward and the symmetry axis has three positions:

① On the left side of -2 ② On the right side of 2 ③ Between -2 and 2.

Combined with δ > 0, it can be seen that none of the three cases can be established. Symmetry axis x=(k? -1)/2k. Therefore: ① x=(k? - 1)/2k 0。

②x=(k？ -1)/2k > 2 and f (2) > 0. The solution is -2+√ 5 < k < 2.

To sum up, we can get: -2+√ 5 < k < 2.