Known set A a 2, a 1 2, a2 3a 3 , if 1 A, find the set of values of the real number a

Because of 1 a, so.

If a+2=1, the solution is a=-1, then the set is, the elements are repeated, so it is not true, i.e., a≠-1

If (a+1)2

1, the solution is a=0 or a=-2, when a=0, the set is, the condition is satisfied, that is, a=0 is true

When a=-2, the set is, the elements are repeated, so it is not true, i.e. a≠-2 If a23a+3=1, the solution is a=-1 or a=-2, and it is known that it is not true so the set of values of the real number a that satisfies the condition is

Solution: Discussed in three situations:

Let a+2=1, and the solution is a=-1

In this case: (a+1) =(-1+1) =0

a +3a+3=(-1) +3· (-1)+3=1=a+2 obtained by the heterogeneity of the set elements a=-1 does not satisfy the topic.

Let (a+1) = 1, and the solution is a=0 or a=-2

When a=0, a+2=0+2=2;a +3a+3=0 +3·0+3=3, the set a=, a=0 satisfies the topic.

a=-2, a+2=-2+2=0,a +3a+3=(-2) +3· (-2)+3=1, and a=-2 is obtained from the heterogeneity of the set elements, which does not satisfy the topic.

Let a +3a +3 = 1

a²+3a+2=0

a+1)(a+2)=0

a=-1 or a=-2, from the above problem-solving process, the two solutions do not meet the meaning of the problem, and are discarded.

To sum up, only a=0 satisfies the topic.

The set of values of the real number a is.

Because of 1 a, so.

If a+2=1, the solution is a=-1, then the set is, the elements are repeated, so it is not true, i.e., a≠-1

If (a+1) 2 = 1, the solution is a=0 or a=-2, and when a=0, the set is, and the condition is satisfied, that is, a=0 is true

When a=-2, the set is, the elements are repeated, and the matter is combustion, so it is not true, and the guess is a≠-2 If a 2 +3a+3=1, the solution is a=-1 or a=-2, and it is not true to know that the real number a of the full liquid ant type foot condition is the set of values is

If 3a-2=1, then a=1But a 2 is also 1, and the dissatisfaction is buried in the praise; If 3a-2=3, a=5 3, the full grandson is disturbed; If 3a-2=a 2, a 2-3a+2=0, a=1 or 2, similarly, when a=1, a 2=1, is not satisfied, so bent Kai Zheng a can only be 2To sum up, the set of values of a is as follows:

If a-3=-3, then a=0, then 2a-1=-1, a2 +1=1;

Establish; If 2a-1=-3, then a=-1;

At this time, a-3=-4, a 2 +1=2;

Therefore, Hu Gao was established;

Therefore, the value of the pants is 0,-1

Because a a, a may be one of the 4 elements.

Categorical discussions. When a=1, a 2+a=a+1=1, the uniqueness of the set is not satisfied, and the dust code is gone.

When a=3, Li Nai, satisfied.

When a2+a=a, a=0

At this time, there are two which brothers in the collection 1.Equally abandoned.

In summary, a=3

1∈a；If a+2=1, the stool is eliminated then a=-1, a=, does not satisfy the reciprocity of the set, a≠-1; In front of the bridge.

If (a+1)2

1, then a=0, or -2, a=0, a=; a=-2, a=;

If |a|=1, then a= 1, from the previous a≠-1; When a=1, a= The value of the jujube stupidity of the real number a is: -2,0,1

Because 1 a, when the ear is raised a+2=1, it is positive, a=-1, a=, which is not in line with the topic, and is discarded;

When (a+1)2

At 1, a=0 or a=-2

When a=0, the simple family, a=, is eligible;

When a=-2, a=, unqualified, discarded;

When a23a+3=1, a=-1 or a=-2, rounded.

composite a=0

So the answer is: 0

Because 1 a, when a+2=1, a=-1, a=, does not fit the topic, discard;

When (a+1) 2 = 1, a=0 or a=-2 when a=0, the ear a=, is eligible;

When a=-2, a=, if it is not qualified, the Jane family is discarded;

When a23a+3=1, a=-1 or a=-2, rounded.

composite a=0

So the answer is: positive 0

Test point: intersection and its operation Thematic code stupidity: calculation questions Analysis: Discussion from a b = -3 b, divided into a-3 = -3, 2a - 1 = -3, a2 + 1 = -3, and a2 + 1 = -3, we must pay attention to the mutual heterogeneity of the elements

Answer: Solution: a b=, -3 b, and a2+1≠-3, late burial.

When a-3=-3, a=0, a=, b=, so that a b= contradicts a b=;

When 2a-1=-3, a=-1, it is in accordance with a b=

a=-1 Comments: This question mainly examines the intersection of sets and their operations, and examines the idea of categorical discussion through common elements

If a-3=-3, then a=0, then a={0, bout=1,-3},b={-3,-1,stool width 1}, there are two common elements -3 and 1, the condition is not satisfied, if 2a-1=-3, then a=-1, then a={1,0,-3} code is bright, b={-4,-3,2}, there is only one common element -3, so a=-1 satisfies the condition, if a 2+1=-3, then a has no solution, in summary, a=-1

What is A2? A 2?

If yes, there are the following solutions:

Since a b=, then both a and b should contain 3

Suppose a, a+1=3, then a=2

b = contains 3, and they are different from each other such as trembling filial piety, so it can. Cave slag.

Assuming a 2 = 3, then there can be no element of slag 3 in b.

In summary, a=2