Basic 10 between math problem sets

c is a true subset of b.

is a subset of a First consider b to be an empty set (an empty set is a subset of any set) when m+1>2m-1, b is an empty set solution m<2

When m+1<2m-1, b is a subset of a, and a={x|x<-5 or x>2} so 2m-1<-5 or m+1>2, and 2m-1 m+1 gives m 2

Take the union so m is an arbitrary real number.

8.Didn't read it. But you just need to figure out the definition of the true subset and the subset and you can do it yourself.

9.As with question 7, try to do it yourself, it will help you.

6, a=b, c is really contained in b

7, m 1 or m less than or equal to -2, b may be an empty set, there is m 2, so m is optional.

8，9，a=2，0，2/3

2) m3 (note: like 7b can be an empty set).

6.a (=)b , c (contained in) b

is a subset of a First consider b to be an empty set (an empty set is a subset of any set) when m+1>2m-1, b is an empty set solution m<2

When m+1<2m-1, b is a subset of a, and a={x|x^2+3x-10≥0｝=｛x|x 2 or x -5}

So 2m-1<-5 or m+1>2, and 2m-1 m+1 gives m3 or m-2

Take the union so m is an arbitrary real number.

then there is m3 or m<2

9. p=｛x|x^2-2x-3=0｝=｛x|x=3 or x=-1}, if s is an empty set, then a=0

If s is not empty, then substitute x=3 or x=-1 into ax+2=0 to find a.

2) -3 m 3 of m+1 -2 and 2m-1 5

The relationship between set and set has the relationship between subset and true subset that is, the inclusion relation【Subset】If any element of set A is an element of set B, we say that set A contains set B, or set B contains set A, and we also say that Set A is a subset of Set B.

True subset] If A is a subset of B, and at least one element in B is not part of A, then Set A is called a true subset of Set B.

The intersection and merger mentioned above are the basic operations of sets.

Relationships are: Contains Does not contain.

The arithmetic relationship is as follows: and complement.

There is confusion between whether an element or a set is a relationship of belonging or not belonging, and between a set and a collection of things that is contained or not included.

Contains b, c is really included in b

7.I can't write either, I just read my first year of high school, and I feel that math is so difficult!!

Define the domain: 2x+3-x 2>0, so -10, so 2x+3-x 2>0 solves -1 so the function defines the domain as (-1,3).

2) First, determine the monotonicity of t 2x+3-x 2 in the defined domain, it is easy to increase monotonically in (-1,1], and decrease monotonically in [1,3), because the function y log4(t) increases monotonically in the defined domain, so it can be obtained from the law of "increasing to increase, increasing or decreasing to decrease" of the composite function.

The function increases monotonically on (-1,1), and the monotonically decrements (3) on [1,3) and the increase or decrease from (2) can be determined, and the function obtains a maximum value at x 1, and the maximum value is y log4(2 3 1) log4(4) 1, so the maximum value of y is 1, which is obtained at x 1.

1.The subset is , , and the empty set, and the true subset is , and the empty set;

2.First solve the set a, a=, and b is a true subset of a, i.e., the elements in a are not completely contained in b, therefore.

When b=, k=

2；When b=, k=-(2 3).

Hello landlord, is your topic not complete?

ax²+2x+1=？What is the same But I guess it's equal to 0 first question. There is only one element discussed in 2 cases :

is a one-time function that has only one intersection with the x-axis.

Discriminant =0 i.e. 4-4a=0 a=1

So the answer to the first question is a {0,1}

Second question. At least one element.

It can be understood as having one or more elements It is also discussed in two situations:

Discriminant =0 i.e. 4-4a=0 a=1

So the answer to the second question is a 1

A set of 15 people love basketball.

Set of 10 people love table tennis.

A friend loves basketball and table tennis.

There are 30-8=22 people who love basketball or table tennis, so there are 15+10-22=3 people who love basketball and table tennis, so the number of people who love basketball but don't like table tennis is 15-3=12 people, you can draw pictures to help understand.

A: The number of people who love basketball but not table tennis is 12.

As you can see from the title, there are 3 people who love basketball and table tennis. So there are 12 people who only love basketball, and there are 7 people who only love table tennis, so there are 12 people who love basketball but don't like table tennis.

A intersection b represents x 2-ax-a 2-19=0 and x 2-5x+6=0a and b represents the total set of solutions covered by x 2-ax-a 2-19=0 and x 2-5x+6=0.

Since the set of solutions for b is (2,3), then aub has at least 2 solutions, because a intersects b=a and b, then a intersects b and has at most 2 SOLUTIONS, AND BECAUSE A HAS AT LEAST 2 SOLUTIONS, A AND B HAVE THE SAME SOLUTION, I.E., A=B, SO A=5

a b is contained in a, a is contained in a b a b, so a a b, a b = a b is included in a, a is contained in a b, so a a b is the same as b a b, so a b

It's easier to understand with a Venn diagram).

b=, a= x=3, and 2 substitutions give a=5

Since equations A and B are both a one-dimensional quadratic equation, and B has two different solutions, and A B=A B, so A and B must be the same solution to the equation, and the solution of B is 2 and 3, so A must be 2 and 3, and it can be seen that A=5 can be seen through the contrast coefficient, or you can find A=5 by bringing the solution in, I hope it will help you!

It is known by b=

Then a=b=brings 2 into the equation a, and we get a=5 -3;

Bringing 3 into the equation in a, we get a=5 -2;

A should be de-common in order to be satisfied, a=b, so a=5

∵a∩b=a∪b

Set A is the same set as set B.

Solution x -5x+6=0 x1=3 x2=2

Bring the equation into the set a to calculate.

a=x +3x+10=(x+3 2) +31 4 0 always establish a=r so that you don't need to count b, a is the whole set of real numbers, then b must be a subset of a, if you need to calculate b, you can divide it into two cases: non-empty set.

Then when calculating b, it is necessary to divide the situation.

1) b is (empty set), constant holds.

m+1>2m-1

m<2(2)b is not an empty set, anyway, it is always established, so the range of the value of m is r (the whole set of real numbers).