Two math analysis questions, related to the set, please help you!

It doesn't hurt to let b and a not intersect (otherwise the difference b a is considered). Since a is an infinite set, a has a countable subset c = . Let B construct a one-to-one mapping to AUB as follows:

f(c_2k) = c_k

f(c_2k-1) = b_k

f(x) = x, when x does not belong to c

I can't guarantee that the second question will be correct. Give it a try.

For any set A, the elements in A are linear spaces, and in a similar finite case, it is possible to define the straight sum of all linear spaces in A (I can't write this strictly yet), i.e., if the elements in A are linear spaces, then A can correspond to the straight sum of a linear space (or linear space).

In this way, a set of linear spaces establishes a one-to-one correspondence with a linear space.

If t contains all linear spaces, then the power set of t (i.e., the set of all subsets of t) also corresponds to a set of linear spaces. By the maximumness of t, the power set of t corresponds to a subset of t. It's impossible, contradictory.

The only problem with my proof is how to define the straight sum of linear spaces, which I think can be defined, but I don't know how to write it rigorously (probably using the axiom of choice?). ）

1. B belongs to A, so A and B = A proves that the absolute value of A = the absolute value of A and B.

1.A and B are traveling in opposite directions next to the railway, the speed is 1 meter per second, a train is heading towards A at an equal speed, the train has taken 16 seconds to drive by A's side, and it has taken 18 seconds to drive by B's side, how many meters is the train long?

Suppose the train speed is x meters and seconds, then.

16*(x+1)=18*(x-1)

16x+16=18x-18

16+18=18x-16x

34=2xx=17

Train length = 16 * (17 + 1) = 16 * 18 = 288 meters.

2.Passenger cars and trucks drive out of the two cities at the same time, and the buses travel 60 kilometers per hour. Trucks travel 48 kilometers per hour. The two cars met at a distance of 18 kilometers from the midpoint, and the distance between the two cities A and B was found.

Assuming that the distance between the two cities is x kilometers, then.

1 2*x+18) 60=(1 2*x-18) 48 denominator, multiply 240 on both sides of the equation at the same time, there is.

4*(x/2+18)=5*(x/2-18)2x+72=5x/2-90

72+90=5x/2-2x

162=x/2

x = 324 (km).

3.The two sisters set off from home to school at the same time, the sister walked 80 meters per minute, the younger sister walked 60 meters per minute, and after 3 minutes, the sister found that she forgot to bring her language book, and immediately returned to pick up the book and went to school, and finally arrived at the school at the same time with her sister, how many meters from home to school?

Let's say both of them took x minutes to get to school, then.

80*(x-3*2)=60*x

80x-480=60x

80x-60x=480

20x=480

x = 24 (minutes).

The distance from home to school is 60*24=1440 meters.

Hope it helps.

If you have any questions, you can follow up.

Thank you.

The probability of the first occurrence of 15 is 1 49

The probability of the second occurrence of 15 is c(48,5)*c(44,1)=1 75341376,, the probability is very low, so it is different.

The first is 1 49 possible, and the second requires all combinations of 6 numbers from 1 to 49, and then divides by 49 numbers to produce 6 numbers containing 15 probability. It should be different, the combination alone is enough of a headache, and it has to be proven! Hehe!

I can't fix your question, I can't read your question.

<> you have made half of the rent, and the hall-shaped rock dress is turned into polar coordinates.

It can be seen from the question.

The presence of point c allows f(x) to be maximized.

Knowable f'(c)=0

Know f'(a)=f'(b)=0

So in the range (a,c) and (c,b) there are points d, and f is f'(x) Obtain extreme values.

to get f''(d)=f''(f)=0

In the same way there is in (d,f), but g makes f''(x) Obtain the extreme value i.e. f'''(g)=0

So there is at least one heel.