A a 2, a 1 squared, a squared 3a 3 , 1 belongs to A, find the value of a

The key to solving the problem is the uniqueness of the elements in the set.

1. If a+2=1, then a=-1, (a+1) 2=0, a 2+3a+3=1, there are repetitive elements, so a=-1 does not satisfy the topic.

2. If (A+1) 2=1, then A=0 or -2, A=0, A+2=2, A2+3A+3=3, there are no repeating elements, so A=0 satisfies the topic.

When a=-2, a+2=0, a2+3a+3=1, there are repeating elements, so a=-2 does not meet the topic.

3. If a 2 + 3a + 3 = 1, then a = -1 or a = -2, as can be seen from the above conclusions, are not in line with the topic.

In summary, a=0

It's sorted. 1) a+2-1, a==1 (a+1) 2=0 a 2+3a+3=1 There are 2 elements that are the same and do not match.

2)(a+1) 2=1 a=0 or -2 equals 0 a+2=2 a 2+3a+3=3 fuhe

When equal to -2, a+2=0 a 2+3a+3=1 fuhe3)a 2+3a+3=1 a=-1 or -2 is equal to -1 a+2=1 There are 2 elements that are the same and do not match.

When equal to -2 (a+1) 2=1 there are 2 elements of the same non-conformity.

The composite is a = 0 or -2

My oh. I'm a master, and I can't go wrong with that.

From the equation, 1 = 3a-a squared, so 1 a = 3-a, so (a-1 a) peaceful and enlightened square = (2a-3) square next to = 4a square to call dust -12a + 9 = -4 + 9 = 5

a=0Original set = {-3,-1,-4} is true.

a=-1Original set = {-4,-3,-3} Does not satisfy the heterogeneity of the set elements Leave off the early lead.

squared -4=-3 a=1 or a=-1

When a=1 the original set={-2,1,-3} holds.

When a=-1 the original set is closed = {-4,-3,-3} the sedan does not satisfy the heterogeneity of the set elements.

In summary: a=0 or a=1

If a-1=-2, then a=-1

then the condition a= has duplicate rounding.

In the same way: 2a 2 + 5a + 1 = -1, solution a

a 2 + 1 = -2, solution a

Then bring in the expression in the set a, and then the mutually heterogeneous song orange that conforms to the set becomes the field brigade group.

Solution: a+1 a=3

It can be obtained on both sides.

a²+2+1/a²=9

So a +1 a = 9-2 = 7

Because the mu section is a 2 1>=1, the burning is lifted.

A 1 2, a 1, Xun Qian because 2a 2 + 5a + 1 = -2 at this time, is not established, and is discarded.

2a 2 + 5a + 1 = -2, of a

Solution: According to the question: a-1=-2 or, the round manuscript disturbs the orange dan2a*2+5a+1=-2or a*2+1=-2 (to discard) the shirt, so a=-1 or a=-1 2 or a=-3

If a-1=-2, then a=-1, then 2a squared + 5a+1 is equal to -2, upshifting and violating the principle of mutual difference of set elements. Therefore, only 2a square + 5a + 1 can be noisy equal to -2.

List the equations and solve them a=

or a=-1, so a=

Two scenarios:

a-1=-2.

a = -12a square + 5a + 1 = -3

A squared + 1 = 2

a=satisfying the omission.

The collection of state search cavities is heterogeneous.

2a square + 5a + 1 = -2.

a=-1 or a=-3 2

a=-1.

a=Not satisfied.

The heterogeneity of the set.

Abandon it. 2): A=-3 at 2. a=

Because the shouting field is: a 2--3a+1=0

Obviously, a≠0, so both sides are divided by a at the same time, and we get:

a--3+1 Zheng shouts a=0

So: 1, a+1 wax skin a=3

2、a^2+1/a^2=(a+1/a)^2--23、(a--1/a)^2=a^2+1/a^2--2=5