Take a look at this to the set of problems high school math .

The correct one should be like this, you can refer to the comparison, and you will know that it is not right.

1.When xy=0, there are three cases, x=y=0 or x=0,y≠0, or x≠0,y=0

When x=y=0.

x+y=x-y=0 does not conform to mutual heterogeneity.

So give it up.

When x=0,y≠0.

x-y= x -y and x+y=x +y gives y=1, x=0x-y= x +y and x+y=x -y gives y=-1, x=0 when x≠0, y=0.

x-y= x -y and x+y=x +y gives y=0, x=1x-y= x +y and x+y=x -y gives y=0, x=1x-y=x+y does not meet the mutuality rounding.

2.When xy ≠ 0, i.e., x≠0, y≠0

x-y= x -y and x+y=x +y gives y=1, x=1x-y= x +y and x+y=x -y gives y=-1, x=1x=1, and x -y =0 when y= 1

Abandon it. In summary, when x=0, y=1, p=, q=x=0, and y=-1, p=, q=

Your first calculation at xy=0 is wrong, xy=0, but the probability of your solution is x=1, y=1 or x=0, y=0, which is obviously wrong, that is, you ignore this condition when you assume xy=0

It should be: x=0, y=1 or x=0 y=-1 or x=1, y=0 or x=-1, y=0

The retest yields x=0, y=1 or y=-1

p=q=

Disrupted, ignoring the main premise.

1) A b= , what is the relationship between the line l and the circle o.

2) When a b is a single element set, what is the relationship between the line l and the circle o tangent 3) when a b is the set, the line l and the circle o intersect 2 element levels.

2.Knowing that x>0,y>0, x+y=8, then the maximum value of xy is 163Given that x>0,y>0,xy=9, then the minimum value of x+y is 6

Proof that: (1) If there is x s2 for any x s1, i.e., s1 is contained in s2, then s1 s2=s2 is a true subset of r, so there must be c r so that c s1 s2

2) If s2 is contained in s1, the same can be seen that there is c r so that c s1 s2

3) If neither of the above two cases is true, i.e., there is x s1 so x s2, and there is y s2 so y s1, then consider x+y r, if it belongs to s1, then there should be y=(x+y)-x belongs to s1, contradiction!Therefore it does not belong to s1. In the same way, it does not belong to s2, so x+y s1 s2 is proven.

Select a because m= and set x1, x2 m

Let x1=a+ 2b, x2=c+ 2d, then x1+x2=(a+b)+2(c+d), where =(a+b)(c+d) z, so option A is correct!

The first few "weird numbers" need to be tried, and it should be noted that 1 is neither prime nor composite. It is easy to see that 9, 11, and 13 are all "weird numbers". In fact.

For 9:2,3,5,7 are prime numbers;4, 6, 8, 9 are composite numbers.

For 11: 2, 3, 5, 7, 11 are prime numbers;4, 6, 8, 9, 10 are composite numbers.

For 13: 2, 3, 5, 7, 11, 13 are prime numbers;4, 6, 8, 9, 10, 12 are composite numbers.

Let's illustrate that there are only these three "weird numbers", i.e., a=.

Counting down from 13 is 14, 15, 16, all three are composite. So by 16, there are three more composite numbers than prime numbers. Every time a prime number is counted in the future, its next number is an even number, and it must be a composite number.

Therefore, starting from 17, for every positive integer n, the number of composite numbers is at least two more than the number of prime numbers, i.e., it is impossible to be equal. So 9, 11, and 13 are all "weird numbers".

Oh yes, 1 yes too, I ignored it. Because a positive integer that does not exceed 1 has only 1 itself, and the number of prime and composite numbers is 0, 1 is also a "weird number".

cua=,k∈z}

cub=,k∈z}

So there is: a is the true subset of cub and b is the true subset of cua.

1. If a belongs to a, then 1 1+a belongs to a, then 2 belongs to a, 2 satisfies this formula, bring in, and get that 1 1+a is equal to 1 3, and in the same way 1 3 also belongs to a, and also satisfies this formula, bring in, and get that 1 1 + a is equal to 3 4

2. To make a a set of single elements, a needs to be equal to 1 1 + a, assuming that it is equal to, dissolve, a2+a-1=0, and the solution is obtained.

a=（-1±√5）/2.

1. Because if a belongs to a and a is not equal to 1, then 1 1+a belongs to a (1 (1-a)), otherwise the subsequent calculations cannot be continued, and a≠1 is meaningless. I'm here according to 1 (1-a).

and a = 2, then 1 (1-a) = -1 a

a=-1, then 1 (1-a)=1 2 a

a=1 2, then 1 (1-a)=2 a

So the other two elements are -1, 1 2

2. If a is a unit set.

then a=1 (1-a).

So a -a+1=0 and there is no solution to this equation.

So a can't be a single element set.

Solution:1∵2∈a

2≠1∴1/(1-2)=-1∈a

-1≠11 [1-(-1)]=1 2 a

1 2≠1

1 (1-1 2) = 2 belongs to a

The other 2 elements in a are -1 and 1 2

2.If a is a cell set, i.e. a has only one element.

a=∵a=a=1/(1+a)

The solution is a=....Do the math yourself)

I think it's probably not that you don't know how to do this kind of question, but that you can't understand it.

I'll give you a translation of the title!

The set a is actually a = its elements are the solution of this equation.

The set b is actually b=, and its elements are the solutions to this equation.

Then look below.

a=b≠ , indicating that the solutions of both equations are the same, and that they both have solutions.

Then it's time for junior high school questions.

The range of values of a is discussed.

a:(x-2)(x-1)=0 so a={1,2}

1) Because a b = {2} so x = 2 is a solution of b, substituting it in: 4 + 4 (a + 1) + a2-5 = 0

Therefore, a2+4a+3=0 is solved to obtain a=-1 or a=-3 and both are consistent after testing.

2) Since a b = a, when b is an empty set, 0, the solution is a -3

When b is not an empty set is, a -3, to make a b = a, then b can only be or or since or means b has only one root, so when b is or, it should be = 0, i.e. a=-3, and when a = -3, the solution is x=2, so a = -3 is conditional.

How to use the Vida formula when b = to know that there is no such a

To sum up: a -3

3) To make a (cub)=a, a b must be an empty set.

When b is an empty set, the condition is met, so a -3 is eligible.

From (2), it is known that the set of b cannot have this solution, and when the solution of b contains 1, we get a = 3-1, or a = - 3-1

When the solution of b contains the solution of 2, we get x=-1 or x=-3

Therefore, the range of values of a is that a cannot be equal to - 3-1, a = 3-1, x = -1, x = -3

(1)a=.Because a b = {2}, so 2 belongs to b and substitute x = 2 into x + 2 (a + 1) x + a -5 = 0 to solve a = -1 or a = -3, and test a = -1 or a = -3, which is correct.

2) Because a b = a, then b = empty set, b =, b = when b is an empty set, discriminant = 4 (a + 1) 2-4 (a 2-5) < 0, so a <-3

When b=, a=-3, a=-1 (rounded).

When b = or a, it doesn't fit the topic or has no solution.

So a<=-3

3) A (cub) = a, so a b = empty set substituting x = 1, x = 2 into x + 2 (a + 1) x + a -5≠ 0 to get a≠ -1 + root number 3, a≠ -1 - root number 3, a≠ -1, a≠ -3