Please help me solve a math problem!

I'll give you the answer? Or a hint!

Question 1, I can't figure it out like that, I'll break it down one by one, the first one is 1-1 2, the second one is 1 2-1 3 in turn. The last one is 1 2009-1 2010 The answer comes out right away: 1-1 2010 = 2009 2010

Question 2, which looks the same, is to multiply all the numbers by 6 and divide the result by 6.

It becomes (1 10-1 100) 6 = 3 200

Please take a closer look:

Because 1 1 2 1 1 2 1 2 1 2, 1 2 3 1 2 1 3 1 6, as above, after removing the brackets, the front and back are offset to get 1 1 2010).

In fact, it is similar to the above question, except that the denominator is a few numbers different, and it is multiplied by a fraction at the end, such as (1 3 5) (1 3 1 5) 1 (5 3) 1 15).

There is a pattern to this kind of question, and the result of this question is:

1 1 1 1 n

1 2 2 3 3 4 n (n+1) n+1 maximum - minimum.

The result of this question is: -

Minimum Number Maximum Number The difference between the two numbers of the denominator.

Cost of hail earned = 12 (1+20%) = $10.

The cost of loss = 12 Qingzhou sail (1-20%) = 15 yuan.

12+12-10-15=-1 yuan.

If one of the two good clothes is sold, you will lose 1 yuan.

Lose money, earn 20%, the cost is 10 yuan, earn 2 yuan, lose 20%, then the cost of selling the sail is 15 yuan, lose 3 yuan, so each sold 1 piece and lost 1 yuan.

c = 3÷60×2πr

cm) A: The tip of the needle is walked over the centimeter.

Suppose the two places are x kilometers apart, and car A travels X+48 and car B travels X+X-48 when they meet for the second time, and the distance ratio traveled is the speed ratio because the time is the same.

x+48):(x+x-48)=5:4 x=72

Let the distance between A and B be s, the velocity of A is 5V, and the velocity of B is 4V, which is obtained according to the equal time taken by AB.

s+48)/5v=(2s-48)/4v

This gives us s=72

The two places are located at a distance of x kilometers.

Car A has traveled X+48 km, and car B has traveled X+X-48 km.

Because the time is the same. So the distance traveled by car A 5 = the distance traveled by car B 4 is calculated as x = 72 kilometers.

Let the distance be x km, then (x+48) 5=(x+x-48) 4, x=72 km.

Turns out big x turns out small x-30

Later large x+5 : Later small x-35 = 7:4

x is not an integer, is the problem wrong?

(1) First, the right side of the equation is divided into a(x-3), b(x+1) (x+1)(x-3), and according to the identity property, a(x-3), b(x+1), x+5 can be obtained

According to this equation, the system of equations a-b=1 and -3a b=5 can be listed and solved to obtain a=-1, b=-2

2) The average unit price, the cost of two purchases and the quantity of two purchases. Let the first unit price be A, and the second unit price be B.

The average unit price of A is 1000a+1000b, 1000+1000, simplified to A+b 2

The average unit price of B is 800+800 (800 a+800 b), which is simplified to 2ab a+b

To judge whose purchase method is more cost-effective is to see whose average unit price is low, with the unit price of A and the unit price of B, if "0, then the unit price of B is low.

If < 0, the unit price of A is low. (a+b 2) (2ab a+b) is simplified after the division to obtain (a-b) 2 2(a+b) because according to the actual a>0, b>0, so the original formula is also "0, so the unit price of B is low, and B's purchase method is more cost-effective.

1.Let's say a total of x eggs are laid.

Then the first hatch is less than two-thirds of five, that is, 2x 3 -5 eggs, and the second time the remaining one-half is less than 3 eggs, that is, (x-(2x 3 -5))*1 2)-3=x 6-1 2 eggs.

Since the number of eggs is a natural number.

So the first hatch of 2x 3 -5>0==>x> and x is divisible by 3, and the second hatch x 6-1 2>0===>x<12 combined min.