Junior high school math problems, prawns help

1) Diagram can be seen as 1 4, diagram can be seen as 2 4, diagram can be seen as 3 4, diagram can be seen as 4 4, diagram can be seen as 5 4, diagram can be seen as 6 4;

2) According to the law of (1), the graph n can be regarded as 4n, and the solution is solved according to the formula of summing the natural numbers: according to the figure in the problem, it can be:

The first body shape requires 1 4 = 4 pieces;

The second body shape requires 4+4=2 4=8 pieces;

The third body shape requires 4+4+4=3 4=12 pieces;

The fourth body shape requires 4+4+4+4=4 4=16 pieces;

The fifth body shape requires 4+4+4+4+4=5 4=20 pieces;

The sixth body shape requires 4+4+4+4+4+4=6 4=24 pieces;

The nth body shape needs 4+4+4+....+4=4n pieces How many pieces does it take to place the 20th square in this way?

The 20th body shape needs 4+4+4+....+4 = 4 * 20 = 80 pieces.

Your first one is a point ...

From the second point onwards, the number is the point (n) on one side (n) minus the common point 4.

The first 1 (does not conform to the law below), the second 4=2 4 -4, and the third 8=3 4-4

The nth 4n-4

20th = 4 20-4 = 76

Let me start with the first question, the first square needs 2 chess pieces for each side, the four sides are 2*4=8, and then subtract the four overlapping chess pieces to get 8-4=4, in the same way, the nth square needs n+1 chess pieces for each side, and the four sides are 4*(n+1), and then subtract 4, that is, 4*(n+1)-4=4n. If you don't believe it, bring in the second square and try it, it's 8...

I don't need to talk about the second question.

1, n squares to 4n.

Each requires 80 pieces.

Solution:1For example, the first one asks for 4 flags. The second is 8The third 12, the fourth 16The nth is 4n.

It's simple: 4n (n is the first few graphs).

The second question is 4 20 = 80.

How are you? We haven't been in touch for a long time, is everything okay.

1. When A typed 500 8=4000 words, B typed 600 7=4200 words, that is, for every 4000 words A typed, B typed 200 words more than A; A typed two pages and typed 1000 words, so B wanted to catch up with A, and A typed (1000 200) 4000=20000 words, that is, 40 pages.

2.The ratio of the velocity of A, B and C is 100:(100-5):(100-10)=20:19:18

Therefore, when B reaches the end point, C can only run 100 * 18 19 = meters in the same time, so C still has meters at the end of the line.

1.When A typed 500 8=4000 words, B typed 600 7=4200 words, that is, when A typed 4000 words, B typed 200 more words than A, then B typed 200 8=25 words more than A per page, so the number of words A and B to type is the same, then B needs to type 2*500=1000 words, because B typed 25 more words per page, then B has to type 1000 25=40 pages to be the same as A;

2.The ratio of the speed of B and C is (100-5) :(100-10) = 19::18, so when B reaches the end point, C can only run 100 * 18 19 = meters in the same time, so C still has meters to the end point.

1. It's 10 pages, because A has already typed 1000 words, so B only needs to catch up with A's 1000 words, and B has 100 more words than A every time, so B only needs to write 10 pages, and it's the same!

2. Of course, it's 5 meters, because they all move in a straight line at a uniform speed!

This is a kind of problem, which can be boiled down to physics, different speeds, different departure times, and the same ...... distanceI'll give you QQ to say it......

MS is more difficult to calculate.

Let 2008x =2009y =2010z =k (2008)x= (2009)y= (2010)z= (k), let =a

That is, left=a x+a y+a z=a(1 x+1 y+1 z) writes the left form as (bk), i.e. (b) k)=a (b)=a(1 x+1 y+1 z).

So 1 x+1 y+1 z= (b).

This calculation is (2008 2009 2009 + 2007 2009 2010 + 2008 2008 2010) (2008 2009 2010)), the latter is incompetent and can not be simplified, but I remember that similar questions seem to be able to be turned into a very simple formula.

Let 2008x = 2009y = 2010z = k

Respectively, it means that xyz will come out soon.

In fact, replace the right side of the equation with a simple symbol, and then you can simply find x, y, z, and then select the common denominator as 2008*2009*2010 through the general divide, and the result will come out. I won't give the answer. Very simple.

3*2008 3*2008 3*2009 3rd Root 2009 I wrote, not necessarily right.

If you want the process, turn on the voice.

Don't want to type...

Solution: Change 2 to x=2y+1 and bring 1 to get :

y -10y+9=0=(y-1)(y-9) is solved: x=5 when y=1

y=9 x=19

Multiply the formula by 2 to get 2x-4y-2=0, subtract this formula with the first formula, sort out and get y-10y+9=0, that is, (y-1)*(y-9)=0, y1=1 or y2=9, substituting the formula, solving x1=3 or x2=19