To solve the senior one math set problem, please explain the senior one math set problem

11, a greater than or equal to 2

12, just put -1/3 into it.

1. First bring -1 3 to the two equations of the set to calculate the values of p and q (the reason why -1 3 is brought to the equation is because -1 3 is the value of a intersecting b, that is, the x of both a and b equations has a value of -1 3); p = minus 14, q = 3

2. Then bring p and q in, find the knots of the two equations respectively, write out the values obtained, and remove the repetition, which is the result.

By substituting x=-1 3 and finding both p and q, the solution of x of the two equations will naturally come out, and the set of a and b can also be found, and naturally a and b will be found.

Solution: a b= -1 3 is the common solution of a, b, and it can be obtained by bringing -1 3 into the original equation, 1 9 3 + p(-1 3)—5=0, and the solution is p=-14

1 3-10 3+q=0, q=3

Bring p,q into the original equation and solve a=,b==,a b==

11 questions a>=2

Question 12 Bring -1 3 into two equations to calculate p=-14, q=3. Then find the solution set of the two formulas and find the union, and the final result is -3

Basically forgot about it.

Search for my name.

Substituting -1 3 into the two set equations of a and b is obtained to obtain p and q; Resolve the set.

Solution: Set p==

Since p intersects m=

So m=so m and p=

So, there are 15 empty sets of the true subset of set s ,,,.

Since there are heroes in the first question, let me answer your other questions. Since you're in the prep course, I'll try to explain it in as much detail as possible.

1) The basic characteristics of sets: deterministic, different, disordered. From the certainty, x 2-5x+9 should be = 3, and the next thing is to solve the problem of a quadratic equation, I believe you can complete it.

2) This problem is actually a binary equation solved by two conditions. From b is true to a know, x 2-5x+9 should be = 3, which can be obtained from (1) the value of x, and from 2 belongs to b know, x 2 + ax + a should be = 2(Note:.)

There are two solutions to solve a quadratic equation, and the appropriate trade-offs are made, I didn't do this, you can do it) In this way, you can get the answer.

3) Since b=c, the elements are the same in both sets. Yes: x 2 + ax + a = 1, x 2 + (a + 1) x - 3 = 3

Solve the first equation first, solve x=-1 or 1-a, and then bring in equation 2 to get two sets of answers. Try it yourself!

Another question. It is easy to learn the study of the set, as long as you pay attention to it and figure out the relationship. Personal learning experience is introduced for reference.

In this part of the study, first of all, we must firmly grasp the concept and understand the meaning of some nouns and symbols, which is the premise of being able to do the questions. In the study, it is recommended to draw more schematic diagrams to help clarify the relationship, especially in the learning of "handover" and "merge". In short, think more about the summary, and you will find that the collection is just a piece of cake.

A is included in B then m -3

The final method of this kind of problem is to draw a number line and solve the problem intuitively through graphics, 4, x 2-5x+9}

Make a=x2-5x+9=3 x=2 or 3

4，x^2-5x+9} b=

2 belongs to b, and b is really contained in a

x^2-5x+9=3……Type 1.

x^2+ax+a=2……Type 2.

Consecutive 1 2 gives x=2 a=-2 3 or x=3 a=-7 4c=

b=cx^2+ax+a=1……Type 1.

x^2+(a+1)x-3=3……Type 2.

The solution yields x=3 a=-2 or x=-1 a=-6

Because a={x|-3≤x≤4｝，b=｛x|x m}, and a is included in b

So m -3

Like you, I'm also preparing, and my method is:

Sorting out the knowledge points in a book, to sum up, I think high school is messy, it's all concepts.

Do more questions and find out the shortcomings by doing the questions.

Hello: x -5x+9 is a constant, and it appears as an element in a set.

1)∵2∈b

x²+ax+a=2

and b is really contained in a

x -5x+9=3 x=2 or x=3

Substitute x + ax + a = 2 a = -2 3 or a = -4 7 i.e. x = 2, a = -2 3 or x = 3, a = -4 7 (2) b = c

x + (a + 1) x-3 = 3 and x + ax + a = 1 gives x = 3 and a = -2

x=1，a=0

If you have any questions, you can continue to add them, and it is recommended to check them yourself after taking the answers. I could have miscalculated.

(1) 2 b, can only be x + ax + a = 2 .

b is really contained in a and 3 b, which means 3 a, so x -5x+9=3 is solved to x=2 or x=3

If x=2, substitute a=-2 3

If x=3, substituting a=-7 4 yields a=-7 4

So x=2,a=-2 3, or x=3,a=-7 4(2)b=c, illustrates x +ax+a=1 .x²+（a+1）x-3=3...Get x=a+5 substitution to get a=-2 or a=-6, so x=3 or -1

Termination: x=3, a=-2 or x=-1 a=-6

(1) 2 b, which means x2+ax+a=2, b is a true subset of a, then 3 is in a, so x2-5x+9=3

Solved by x2-5x+6=0, x=2 or x=3

A=-2 3 when x=2, a=-7 4(2)b=c, then x2+(a+1)x-3=3, x2+ax+a=1 are solved: a=-3 when a=-2, and x=-1 when a=-6

Solution: Since 2 belongs to b, then x*x+ax+a=2, and b is really contained in a, so 3 belongs to a, that is, x*x-5x+9=3 The solution gives x=2 or 3, when x=2, a=-2 3, and when x=3, a=-7 4

2. B=C, then X*X+AX+A=1 and X*X+(A+1)X-3=3 Combine these two formulas, subtract the first formula from the second formula to get X=A+5, bring back the formula to get A=-2 or -6, when A=-2, X=3, when A=-6, X=-1

The answers, the methods are guaranteed to be correct. Give it a good review!!

In the first question, x -5x+9=3 x +ax+a=2 is solved to x=2 a=-2 3 or x=3 a=-7 4

x -5x+9=3 x +ax+a=4 The solution yields x=2 a=0 or x=3 a=-5 4

The second question is x +ax+a=1 and x +(a+1)x-3=3 gives a=-2 x=3 or a=-3 x=-1

a∈b?How could a collection have this?

1. If a belongs to a, then 1 1+a belongs to a, then 2 belongs to a, 2 satisfies this formula, brings in, and obtains 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, and brings in to 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)

1. Solution: It can be obtained from the question, -3 so the true subset that satisfies the meaning of the question has;

The 6th power of 2 is subtracted by one, which is equal to 32-1=31, so choose b (the number of all subsets of the original set is 32, so the number of true subsets is 31) Attached: If a set contains n elements, then the number of subsets of the set is the nth power of 2.

2, is it (x-3) or (x)-33 under the root number, is it a complement of (a b) = ?