Simple high school math compulsory one topic set , come quickly, masters

1.Because , b contains a, so the range of b is less than or equal to a

So. m+1>=-2

2m-1<=5

-3<=m<=3

2.Solution: a b = indicates that there is an element in b that is -3

Let's start the discussion.

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

Substituting a, B gets:

a=,b=at this time a b=

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

Substituting a, B gets:

a=,b=ab=yes.

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

3.From the meaning of the title: b=, c=

1) When a b = a b, a = or a = b

When a=, the discriminant formula: a 2-4 (a 2-19) < 0, a 2>76 3, i.e., a <-2 (19 3) or a > 2 (19 3).

When a=b, there are: -a=-5 and a2-19=6

Solution: a=5

Combining the above two cases, there are: a<-2 (19 3) or a>2 (19 3) or a=5

2) When true contains a b, a c=, a contains element 3, but elements 2 and -4 do not belong to a

Substituting x=3 into x2-ax+a 2-19=0 yields: 9-3a+a 2-19=0

Solution: a=-2 or a=5

Replace a = -2 back to obtain: x 2 + 2x - 15 = 0, x = 3 or -5;

Replace a = 5 to get: x 2-5x + 6 = 0, x = 3 or x = 2, rounded.

So: a=-2

4.Solution: a (cub)=

A must have elements 1,2, and b does not have 1,2

Because a b ≠

b=2 a=，b=

3 a=，b=

b=,b=，b=

Find the set a=, when b is an empty set, m=o, when the solution in b is 2, m=-1 2, when the solution in b is 3, m=-1 3, so the set of the values of the real number m is m=

m=;The specific process is as follows: the solution set of a is x=2 or x=3;Since A contains B, the possible case for B is: 1:

b is an empty set; The set of 2:b is x=2;

The set of 3:b is x=3;It is derived that m=;

Calculate that set A has 2 and 3, A contains B, and B has 2 or 3 or is an empty set. That is, in mx+1=0, x is equal to 2 or 3 or the value of m when there is no solution, and the answer m is calculated to be minus 1/3, minus 1/2 and 0

First, the set of a is solved with two x=3,2(Empty sets are not considered here, because a set must have a solution).

and a contains b, indicating that a is a subset of b (the genus notes the difference between inclusion and inclusion) {a contains b means that a is a subset of b, but a is included in b, then b is a subset of a}

Bringing the two values in a into b, we get m=-1 3 and -1 2,

1.Solution: a b a b

a b is 2, 3 is the equation x ax a 19 0 two.

Substituting x2 yields: A 5 or 3

Substituting x3 yields: A 5 or 2

a＝52.Solution:

The empty set is really contained in a b

a b≠ empty set.

i.e. 2 or 3 is the root of the equation x ax a 19 0.

again a c empty set.

Neither 2 nor 4 is the root of the equation x ax a 19 0.

In summary: 3 is the root of equation x ax a 19 0, but 2 is not the root of equation x ax a 19 0.

Substituting x2 yields: A 5 or 3

Since x cannot be equal to 2, then a≠5 and a≠ 3

Substituting x3 yields: A 5 or 2

a＝－2

1）a=52) a=0

It doesn't seem right, and it seems to be right, and I'm not sure.

1. Because the condition of (1) states b=a

So substitute b into a, substitute 2 into a=5 or 3, and substitute 3 into a=5 or —1

So a=52 · no, let me tell you back when I get back.

Solution: (1) It's really simple.

a b=a b (you can also draw a Wen En diagram, this will help you understand) a = b substituting 2 and 3, 4-2a+a -19=0 9-3a+a -19=0a -2a-15=0 a -3a-10=0a1=5 a2=-3 a1=5 a2=-2a=52) according to the title, a b ≠ an empty set and a c = an empty set.

There must be element 3 in a, and there are no elements -4 and 2

Substituting 3, 9-3a+a-19=0

a1=5 a2=-2

When a = 5, x = 2 and 3, round off because there is element 2 when a = -2, x = 3 and -5, there is no element -4

So: a=-2

(cua) b = get 2 and get q = 6 in b and b =

cub) a=get 4, in a get p=-7, a=

then a b=

This question mainly looks at who is the element of the set, B is based on y, and C is based on (x, y) as the element.

b = Since n represents the set of natural numbers, both x and y are large or equal to 0Therefore, drawing an image with y=-x +6 only needs to keep the first quadrant. Then x can only take 0,1,2, and the corresponding y can take 6,5,2, so b=

c= The elements represented by this set are the set of points, as explained above. Hence c=

Set of points in the fourth quadrant in the Cartesian coordinate plane Since it represents a set of points, the fourth quadrant is characterized by x>0, y<0 and therefore a=

The set of points on the parabola y=x2—2x+2 Since it represents the set of points, it represents all the points on y=x2—2x+2, so there is d==

1 The following statement is true (

a A set of young people in a village b A set of all small positive numbers.

c Collections {1,2,3,4,5} and {5,4,3,2,1} represent the same collection.

d 1, 0, 5, 1/2, 3/2, 6/4 These numbers form a set of 4 elements.

2 There are four propositions below:

1) whether the smallest number in the set n is; (2) 0 is a natural number;

3) {1,2,3} is a set of natural numbers that are not greater than 3;

4), a n, b n then a + b is not less than 2

The number of correct propositions is ( a, 1, b, 2, c, 3, d, 4.

3 The following four propositions: (1) There is no set in the empty set; (2) an empty set is a true subset of any set;

3) The number of elements in the empty set is zero; (4) There must be two or more subsets of any set

The correct ones are ( A 0 b 1 c 2 d 3 .

The first represents the range of values of the dependent variable of this function, i.e., all the y values in the value range, the second represents the range of the independent variables, i.e., all the x values in the value range, and the third is the ordinal pair, that is, the set of all the points on the function.

So their relationship is:

The first of these is {y|y> or =1}

The second is {x|x belongs to r}

The third is {(x,y)|x belongs to r, y=x 2+1}

y|y=x 2+1} evaluation range.

x|y=x 2+1} to find the defined domain.

x,y)|y=x 2+1} represents a point.

The relationship between the three: the equation is y=x 2+1

The first two are set of numbers, and the third is a set of points.

The first represents the range of the function y=x 2+1.

The second is a defined domain that represents the function y=x 2+1.

The last one is the set of points that represent the points on the image of the quadratic function y=x 2+1, which says that the solution of the quadratic equation y=x 2+1 is the set of solutions.

Analysis:

Let the range of a==y = real numbers greater than 1 =

Let b== be the range of values of the independent variable of y=x 2+1, and let c== be the set of coordinates of all the points of the hyperbola y=x 2+1, and the answer will be easier if you understand what this means.

Answer: C a B (Here.)"≤"is the symbol that represents the subset).

Let's put it this way, let the sets a=, b=, c=, the set a is the domain of the function y=x 2+1, b is the domain of y=x 2+1, and c is the set of coordinates on the curve represented by y=x 2+1 (i.e., the set of coordinates of the definition domain and the value range).

I don't know if I can understand it, and if I don't understand it, I'll bring it up.

y=x 2+1 represents the opening upwards, the lowest point is (0,1) and the axis of symmetry is a parabola on the y-axis.

Then the value of the first set y is (1, positive infinity).

The second x is the whole real number.

The third is the point set.

Therefore, 2 is the largest, then 1, and finally 3

In fact, I can't tell you about the size of the point set and the real number, so don't mislead you! ~