Solution: According to the mutual dissimilarity of sets, a+b= ac, a+2b= ac? Or a+b= ac? And a+2b= ac.

(1) when a+b= ac, a+2b= ac? When,

Then: b=a(c- 1), 2b=a(c? - 1)

There are two kinds: c? - 1=2(c- 1)

Therefore: c= 1

But when c= 1, for set b, there are: a=ac=ac?

So: Give it up.

(2) When a+b= ac? , and a+2b= a,

So: b=a(c? - 1)，2b=a(c- 1)

Two types of division: 2(c? - 1)=c- 1

Therefore: c= 1 or c=- 1/2.

But c= 1

Therefore: c=- 1/2.

2. let A = (x | 1 ≤ x ≤ 3) and b = (x | x ≤ 0 or x≥2), then A∪B is equal to?

Solution: a ∪ b = {x | x < 0 or x≥ 1}