Set A a 2, the square of 2a a, if 3 belongs to A, find the value of the real number a.

a=3 belongs to a, which means that 3 is the element in a

There are two possibilities.

1) A+2=3, then A=1

In this case, since 2a + a = 3, there are two 3s in a set, which do not satisfy the mutual heterogeneity and are rounded.

2) 2a +a=3 then a=1 or a=-3 2

a=1 is known by (1), and is discarded if it does not fit the topic.

a=-3 2, a=appropriate for the topic.

In summary, a=-3 2

If a+2=3 then a=1 then 2a 2+a=3 according to the element is the opposite sex, so a cannot be equal to 1

So there is 2a 2+a=3 but a+2 is not equal to 3 then a=-3 2

Classification discussion: 1) a+2=3, then a=1, at this time, 2a 2+a=3, that is, the set a=, according to the heterogeneity of the set (i.e., a set cannot contain two identical elements), a=this situation is not desirable, so a=1 is rounded;

2) 2a 2 + a = 3, that is, 2a 2 + a - 3 = 0, that is, (a - 1) (2a + 3) = 0, can be solved a = 1 or a = -3 2;

From (1) to know: a=1 is rounded;

a=-3 2, a+2=1 2, at this time, the set a=, satisfies the topic;

In summary: a=-3 2

I hope it can help you, and I wish you progress in your studies!

According to the title. a+2=3

or 2a +a = 3

Solution. a=1

or a=1 or a=-3 2

So the real number a = 1 or -3 2

If a-3=-3, then a=0

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

If the square band of a is -4=-3, then a = 1

Therefore, the value of the real number a can be annihilated by the reed energy of -1,0,1

1), the square of a+2=1a=-1 (side sail a+1) = 0a square + 3a + 3 = 1-3 + 3 = 1 contradiction (2), the square of (a + 1) = 1a = 0 or a = -2 (3) a = 0a + 2 = 2a 2 + 3a + 3 = 3 to establish the fortune hail (4) a = -2a + 2 = 0a 2 + 3a + 3 = 3 = 1 contradiction (blind 5) a square + 3a + 3 = 1a = -2 a = -1a + 2 = 0 (a + 1) 2 = 1 contradiction (6) a = -1a + 2 = 1 (a +1) 2=0...

a b = indicates that there is an element in b that is -3

Discuss. 1) If a-3=-3 gives a=0

a=,b=This finger is only the code stove a b=

Not true. 2) If 2a-1=-3, we get a=-1

a=,b=yes.

To sum up: a=-1 is what is being sought.

a b = indicates that there is an element in b that is -3

Discussion (1) If a-3=-3 gives a=0

a=,b=at this time a b=

If it does not meet (2) if 2a-1=-3 gets a=-1

a=,b= conforms to the above: a=-1 is what is sought.

b = , the square of a + 1≠-3, then.

a-3=-3, a=0, but a crosses b, =, (rounded) or 2a-1=-3, a=-1

At this point a=, b=, a intersects b=, so a=-1

A2 and A2+1 are positive, so it can only be A-3=-3 or 2A-1=-3

When a=0, a=(0,1,-3)b=(-3,-1,1)a, b=(1,-3) does not match.

A=-1A=(1,0,-3)B=(-4,-3,2) A:B=(3).

There are two scenarios:

Lidan hits Chi Liang 1).

A-2=-3, a=-1

2a^2+5a=-3=a-2

This contradicts the definition of a set. (There can be no duplicate elements in the set) so that a is not equal to -1

2a 2+5a=-3, the solution is obtained.

a=-1 (rounded) or.

a=-3/2

test, when a=-3 2, 2a 2+5a is not equal to a-2, so a=-3 2

Either a-2=-3 or 2a squared +5a=-3, both of which cannot be -3 at the same time; The former a=-1, when this repentance is repentant, the flat base spring square of 2a + 5a = -3, then it is rounded, only the square of the Bifeng sedan of 2a +5a=-3, a=rounded)

If 4 set a

then a -a+2 = 4 or 1 - a = 4

When a -a+2 = 4

a²-a-2=0

a-2)(a+1)=0

a=2 or a=-1

When 1-a=4

a=-3A: The value of a is 2 or -1 or -3

a=1. a +a = a + 1 = 2 so it doesn't fit.

a=3. a²+a=12

a+1=4 yes.

When a=a +a.

A=0, where a +a=0

a+1=1, the elements are duplicated, incompatible.

When a=a+1, there is no solution.

In summary, a=3

a=1,3,0

According to the non-repeatable nature of the set, it is obviously wrong to bring in a=0 and get a=(1,3,0,1).

a=1 is brought in, a=(1,3,2,2) is incorrect.

a=3 and a=(1,3,12,4) hold.

In summary. a=3