Substitute x=6 into the formula to get: b=-36-6a.

Substitute x=2 and x=3 into the formula: 0 =

Substitute b=-36-6a into the above two formulas: -9 =

So: a=-9

Substituting a=-9 into b=-36-6a, we get b= 18.

2. Solution:

|x-[(a+ 1)^2]/2|<； =(a- 1)^2/2

【x-(a+ 1)^2/2]^2<； =[(a- 1)^2/2]^2

(x-a)[2x-(a^2+ 1)]<； =0

x 1 = x2=(a^2+ 1)/2

if(a 2+ 1)/2 >； A means that a is not1a = {x | a.

if(a ^ 2+ 1)/2

If (a 2+1)/2 = a, that is, a= 1 A is an empty set.

Therefore, when a= 1 and a = {x | a, a is an empty set.

x^2-3(a+ 1)x+2(3a+ 1)<； =0

(x-2)[x-(3a+ 1)]& lt； =0

x 1=2 x2=3a+ 1

If 3a+1>; 2 is a>1/3b = {x | 2 <; = x & lt=3a+ 1}

If 3a+ 1

If 3a+ 1=2, that is, a= 1/3 B is an empty set.

discuss

When a> 1/3 and a is not equal to 1

A = { x | a & lt= x & lt=(a^2+ 1)/2} 2 & lt； = x & lt=3a+ 1}

a & gt=2

(a^2+ 1)/2<； =3a+ 1

Solution 2

When a< is in 1/3

A = { x | a & lt= x & lt=(a^2+ 1)/2} b = { x | 3a+ 1 & lt； = x & lt=2}

a & gt=3a+ 1

(a^2+ 1)/2<； =2

Solution-radical number 3

When a= 1/3, b is an empty set and a has no solution.

When a= 1, A is an empty set and A is all real numbers.

therefore

The value range of a is [-root number 3,-1/2] u [2,3+root number 5] or a= 1.