The math set problem in the first year of high school, you can understand it

set a=, b=, and a b = empty set, for: (y-3) (x-2)=a+1y=(a+1)(x-2)+3

Bring in b (a 2-1) x + (a-1) y = 15a 2-1) x + (a-1) [(a + 1) (x-2) + 3] = 15 finish: (a 2-1) (2x-2) + 3 (a-1) = 152 (a 2-1) x = 15-3 (a - 1) + 2 (a 2-1) = 16-3a + 2a 2

a b = empty set, then there is no solution to the unary equation of x, and x is meaningless.

So a 2-1 = 0

a=±1

The upstairs methods are all right, but it would be nice if you could imagine the set of 2 points as two straight lines.

A is always a straight line with a slope of (2,3) a+1.

b is a straight line with a slope of -a-1 over (0,15 (a-1)) at a≠1, and b is an empty set when a=1.

Because there are 2 cases that meet the condition, one is that a=1 makes b an empty set;

The second is to make a and b parallel, that is, the slope is equal, a+1=-a-1, and a=-1 is obtained;

The answer is a=1 or -1

But maybe you haven't learned it yet. You can save it and learn it later.

A 2-1 I don't understand this one, is it correct?

I never saw this symbol in my freshman year of high school.

The element x in a is not equal to 2, and a b = empty set, so x=2 in b is messed up next.

First of all, a cannot be an empty set, and when b is an empty set, it is a case, i.e., a=1

When a is not equal to 1, y=(a+1)(x-2)+3 (x≠2), y=15 (a-1)-(a+1) has no solution. That can only be: 3=15 (a-1)-(a+1), and it can be solved.

First, there is one less x in the equation in the b set of lz

You can start by thinking about the reverse side and solving the system of equations:

Eliminate y, and solve x=[6 (a 2-1)]+1 If you want the intersection to be an empty set, as long as this x is meaningless, so a=1 or a=-1

Hehe, if this kind of question is done on the test paper, it is impossible to write the detailed process, and it is done in two steps!

The elements in b are: all real numbers starting from 1 to 4.

When B is really contained in A, that is, all the elements in B are in A, and at least one element in A is not in B, the key in this problem is that since the right endpoint is the same and there are equal signs, then there is no need to look at it, and there is no equal sign in the left endpoint A, and A must contain element 1, so A<1

Because b is really contained in a

So the range of b is less than or equal to the range of a.

So 1>=a

a<=1

B is really contained in A, which means that the scope of B is contained in A.

That is to say, a < 1 cannot be taken, because a=1 means that 1 x 4 is one less value than 1 x 4, that is, the value of x=1, so b is not really contained in a The final result is a<1

b is really contained in a

then a+4<-1 or a>5

So a<-5 or a>5

You don't understand this question very well, but you can solve it in the form of a number line.

b is really contained in the description of a, and b is represented on the number line as being in a, and then you can see it

1. The true subset 、、、 subset is: , , empty set. (Here's a formula:.)

If there are n elements in a set, then its number of sets is 2 of n. If there are two elements in this problem, then the number of its subsets is 2 squared, which is 4. Then the number of true subsets is 4-1=3.

Children's shoes, you can ask the teacher, our teacher emphasized every day in the first year of high school

Is x 1 2x 1=0, yes, a b = empty set, a b = if not, x does not give a range, and the answer to the question of giving a range is different. Also, if you're right, then a b= , a b=r

3、a∩b= 、a∪b=

4、a∩b= 、a∪b=

5、a∩b=

6、cua= 、cub=、(cua)∩（cub).=7, a b={x x=1}, cu(a b)=) I didn't know much about the set at first (now I'm 100% proficient, absolutely correct answer), but if you learn it carefully, you will find it quite interesting. Because it's the easiest chapter of high school Come on!

I'm taking a pen to calculate one question at a time, or just by hand, hopefully! Happy Mid-Autumn Festival! Our Mid-Autumn Festival homework is also a lot of filial piety!

Let's do it together! I've just looked at other answers, and some of them write it incorrectly.

1.The solution is x=4, the subset is an empty set, 4, and the true subset is 4(The symbol of the empty set is not played) 2It doesn't seem to be true, what kind of formula is 2x-1 This really doesn't know how to tell Fu Wang.

Intersect b is 3, and conjuncture b is 0, 1, 2, 3, 4, 5, 6 intersect b is x belonging to 0 to merging b is x belonging to -1 to positive infinity branches.

The intersection b is x=, y= This is a pair of coordinates called an ordered pair of real numbers.

6.The first one is 1, 3, 5The second is 4,5The third balance is 5, and the intersection b is x=the complement of intersection b is x, which belongs to -2 to 1 and goes on 1 to 4 (write greater than or equal to the sign in -2 and 4, less than or equal to the sign in 1, and do not add the equal sign in 1).

1. Subset、、、 true subset

x 1 doesn't understand, is it 2x 1 = 0, if it means yes, a b = empty set, a b =

3、a∩b= 、a∪b=

4、a∩b= 、a∪b=

5、a∩b=

6、cua= 、cub=、(cua)∩（cub).=7, a b={x x=1}, cu(a b)=) These questions are very simple, you should do it yourself in the future, don't leave your studies behind, and then contact if you have something. The QQ number is at the bottom.

It's been a long time since I've done math problems, and I'm a little nervous, and I might be wrong.

1.The solution is x=4, the subset Hezhou is an empty set, 4, and the true subset is 4(The symbol of the empty set is not played).

2.It doesn't seem to be established, what kind of formula is 2x-1, I really don't understand this.

Interchange b is 3, and b is 0, 1, 2, and pants smile 3, 4, 5, 6 intersect b is x belongs to 0 to and b is x belongs to -1 to positive infinity.

The intersection b is x=, y= This is a pair of coordinates called an ordered pair of real numbers.

6.The first one is 1, 3, 5The second is 4,5The third is 5 intersecting b is x = intersecting b and the complement is x belonging to -2 to 1 and on 1 to 4 (write greater than or equal to the sign where -2 and 4, less than or equal to the sign, and no equal sign at 1).

I hope it can help you, after all, I haven't done the question for a while, maybe there is a mistake, forgive me.

1., the true subset is a fierce imitation, [the true subset is all the sets composed of elements that are different from the solution set] 2That set A right?? This is a very tangled topic, we have not done it before--- we have not done it before, and we have to do b=, a and b=

Intersect b = a and b = [this problem can be solved by drawing a number line].

Intersect b = [This question draws two straight line images, and the intersection point is the solution of this problem], cub=, (cua) intersection (cub) = [the last one to see the intersection can directly look at the intersection of the previous two].

Intersecting b = cu (a intersecting b) =

Very simple, junior high school knowledge.

I've only learned this, but I've never understood it in class.

The sophomore of high school has all been returned to the teacher.。。。 Some of the above copy parties are disgusting.

You're a little lazy... Think for yourself first.

Kid: Is that your homework?

The intersection is not an empty set, i.e. two functions have a common point.

y=x-4, so x-4=x-ax+2

x²-(a+1)x+6=0

then the equation has a solution.

Discriminant (a+1) -24>=0

a+1)²>=24

a+1<=-2√6,a+1>=2√6

a<=-1-2√6,a>=-1+2√6

Because 6m+1=3*2m+1, 2m is an even number, and n is an integer in 3n+1, choose c

It is obvious to choose c p=m

Because the k in m is replaced by n+1, it is p

s is their true subset.

A is a straight line (2,3), and B is also a straight line (A≠1), because a b = so a b, or a straight line b passes (2,3), or a = 1 then there is (a + 1) -1 = (a 2-1) (a 2-1) (a 2-1) (a 2-1) or (a 2-1) * 2 + (a - 1) * 3 = 15 to solve a = -4, a = 5 2

So a=1,5 2,-4

m n = empty set.

The set of solutions for a system of simultaneous equations is empty, and this system of equations is solved.

y-3)/(x-2)=a+1

y=(a+1)x-2a+1

Substituting (a 2-1) x + (a - 1) y = 9

a^2-1)x+(a-1)((a+1)x-2a+1)=9(a^2-1)x+(a^2-1)x=(a-1)(2a-1)+9a^2-1=0

a-1)(2a-1)+9≠0

a=1 or a=-1

When x=22a2+3a-14=0

a=, or a=2

The value of a can be (a|).a=1, a=-1,a=2,a=

a+1=y-3 x-2 is brought into (a+1)(a-1)x+(a-1)y=15 to get y-3 x-2(a-1)+(a-1)y=15 According to the problem, if the order equation is unsolved, then only a-1=0, i.e., a=1

a==b=

b contains a, then.

When a=0, it meets the requirements of the topic.

When a>0, (4ac-b) 4a<=3, that is: 4a -3a-1<=0, 0 is synthesized to obtain the value range of the real number a: 0<=a<=1

Find a first, it is not difficult to know a=

Simplifying b, we know y=a*(x-1 a) 2+4a-1 aSince the set of a can be infinite, a>0

Since B contains A, A is a subset of B.

There must be 4a-1 a<=3, and -1 4<=a<=1, so 0

First of all, the value range of y in set a is y>=3

If b contains a, the value range of y in b is greater than a

Let's discuss the case of a=0 first, at this time, y=-2x, the value range is r, when a is not equal to 0, it is a parabola, it must be a>0, and the minimum value must be greater than or equal to the minimum value of a, that is, ax 2-2x+4a>=3, and x belongs to r

After formulation(x-1 a) 2+(4-3 a-1 a 2)>=0, that is, 4a 2-3a-1>=0 and then intersect with a>0, which is a>=1 In summary: a>=1 or a=0

In the set a, y (x 1) 2+3 thus calculates a=. b contains a, that is to say, ax 2 2x 4a in the b set is everstable or equal to 3, then ax 2 2x 4a 3 0 is always established, and then use .

[[[1]]]

The set a is the domain of the function y=x -2x+4.

y=x²-2x+4

x-1)²+3≥3

The value range of this function is [3, +

Set a=[3, +

From the problem design, it can be seen that the value range of the function y=ax -2x+4a contains the interval [3, +when a=0, the value range of the function y=-2x is r, and when the problem is assumed that when a≠0, the combination of numbers and shapes can be seen that there should be a 0, and (16a -4) (4a) 3 solution 0 a 1

To sum up. 0≤a≤1