How to do this Olympiad must have clear ideas and answers. Extra points for a good answer

The work of 4 people in Group A needs to be completed by 5 people in Group B; The work done by 3 people in Group B and 8 people in Group C is required.

This sentence is rewritten as:

The work of 12 people in Group A needs 15 people to be completed in Group B; The work done by 15 people in Group B is completed by 40 people in Group C.

Then find the work efficiency separately and set the workload per person per day to 1

Group A: 1 12, Group B: 1 15, Group C: 1 40

Calculation workload: (10*1 12+20*1 15)*3=So the working time is days.

Ask for a reward for satisfaction o ( o thank you.

If the workload of group A is x, the workload of one person in group B is y, and the workload of one person in group C is z, then the workload of 4 people in group A and 5 people in group B needs to be completed; The workload of 3 people in group B, and 8 people in group C need to complete it".

4x=5y,x=(5/4)y

3y=8z,z=(3/8)y

It is a job that requires 10 people in Group A and 20 people in Group B to work together for 3 days. If the time t=3(10x+20y) 10z is given to 10 people in group C, and y is substituted into the equation as an intermediate variable, t=26 can be obtained

So, if you ask 10 people in Group C to do it, it will take 26 days to complete.

If the total project is "1", then the daily workload of each person in Group A is 1 (12 6), and the daily workload of each person in Group B is 1 (8 10), and the project can be completed in x days.

1 = 4x (12 6) + 5x (8 10) gives x = 144 17 = 8 and 8 17

The project can be completed in more than 8 days.

Digital, refers to numbers.

For example: 13, there are two digits, 123, there are 3 digits with one digit: 1--9, the upper and lower volumes have a total of 9 2 = 18 digits with two digits:

10-99, the upper and lower volumes have a total of 90 2 2 = 360 numbers, and 777-18-360 = 399 numbers remaining.

The rest is the three-digit page number.

There are 7 more pages in the first volume, and (399 + 7 3) 2 = 210 numbers 210 3 = 70

It is the 70th three-digit number, which is 100 + 70-1 = 169 and there are (169) pages in the upper volume.

1 9 pages of 9.

10 99 pages of (99 10 1) 2 180 pages The total number of pages with three digits per page in the two volumes is: 777 (180 9) 2 399

Because the first volume has more pages than the second: 7 3 21.

The total number of pages with 3 digits per page in the next volume is: (399 21) 2 189.

Namely: 189 3 pages 63.

So the next volume is: 99 63 162 pages.

The first volume has 162 7 169 pages.

It's the page number.、Suppose the last book has x pages.,Then the next book is x-7 pages.、Two volumes are x+x-7=777.。。 x is 392 pages.

It's what we call a page number, a number that is marked at the bottom of a book.

Number of pages = (777 + 7) 2 = 392 pages.

If there is something you don't understand, ask again, I wish you to learn and make progress and go to the next level! (

The speed of the ascent is 3 km/h, and the descent speed is 6 km/h. Find the average speed of going up and down the mountain?

Solution 1 has an average velocity of per hour:

2 (1 3 + 1 6) = 4 km.

Solution 2 is a numerical method, and a one-way journey is 6 kilometers.

It takes 6 3 = 2 hours to ascend the mountain.

It takes 6 6 = 1 hour to descend the mountain.

The average round-trip speed is 6 2 (2+1) = 4 kilometers per hour: car A needs 6 hours to travel from A to B, and car B needs to travel 10 hours from B to A. Now car A and car B start from A and B at the same time, go in the opposite direction, car A travels 90 kilometers more than car B when they meet, and the distance between car A and B is found.

Solution 1: The velocity ratio of A and B is 10:6=5:3

AB distance is 90 (5-3) (5+3) = 360 km Solution 2 Two cars meet Required: 1 (1 6 + 1 10) = 15 When 4 hours meet, A has more trips than B for the whole journey

ab distance: 90 1 4 = 360 km.

（1）1+3...19 = 25 * 2 + 50 ,, answer 25 * 2 + 50 + (2) Suppose the average speed of the distance s = s*2 (s 3 + s 6) = 4 (3) assuming the distance A speed a 6 B speed a 10 The speed of A is 5 3 of B, so A walks 5 8a and travels 2 8 = 90 kilometers more than B 3 8a, then the distance is 360 kilometers.

Give points and give points.

, apparently a≠-1

So (a+1)(a 2-a+1)=0

There is a 3+1=0, i.e. a 3=-1

a^2010=(a^3)^670=1

So a 2010+1 a 2010=1+1 1=2

2. (a^2+b^2)^3=(a^3+b^3)^3+8a^3*b^3

a^6+3a^2b^4+3a^4b^2+b^6=a^6+2a^3b^3+b^6+8a^3*b^3

There are 3a 2b 4+3a 4b 2=10a 3*b 3

i.e. 3(a2+b2)=10ab

Divide both sides by ab, and there is a b+b a=10 3

3.Let 2007x 2=2008y2=2009z2=2010w2=t2

1/x=(√2007)/t

1/y=(√2008)/t

1/z=(√2009)/t

1/w=(√2010)/t

The sum of these 4 equations has ( 2007 + 2008 + 2009 + 2010) t=1

i.e. t=(2007+ 2008+ 2009+ 2010).

At the same time, there is. 2007x=t*√2007

2008y=t*√2008

2009z=t*√2009

2010w=t*√2010

The sum of these 4 equations is 2007x+2008y+2009z+2010w=t*( 2007+ 2008+ 2009+ 2010)=t 2

So (2007x+2008y+2009z+2010w)= t 2=t=(2007+ 2008+ 2009+ 2010).

Solution: 1

a^2-a+1=0

a+1/a=1

a^2010+1/a^2010

a^2009+1/a^2009)(a+1/a)-a^2008-1/a^2008

a^2009+1/a^2009-a^2008-1/a^2008

a^2008+1/a^2008)(a+1/a)-a^2008-1/a^2008-a^2007-1/a^2007

(a^2007+1/a^2007)

(a^2006+1/a^2006)(a+1/a)+a^2005+1/a^2005

(a^2005+1/a^2005)(a+1/a)+a^2004+1/a^2004+a^2005+1/a^2005

a^2004+1/a^2004

…Rule: Each step of derivation can reduce the exponent by 3 and change the sign at the same time. The exponent is positive when it is even, and negative when the exponent is odd.

a^2010+1/a^2010

(a^3+1/a^3)

(a+1/a)(a^2-1+1/a^2)

[(a+1/a)^2-3]

2 Incidentally: This problem was solved incorrectly, and it was verified with A 6 + 1 A 6 to know that he was not solved correctly.

2、(a^2+b^2)^3=(a^3+b^3)^2+8a^3b^3

a^6+3a^4b^2+3a^2b^4+b^6=a^6+10a^3b^3+b^6

3a^4b^2+3a^2b^4=10a^3b^3

3(a/b)+3(b/a)=10

a/b)+(b/a)=10/3

3. Let 2007x 2=2008y 2=2009z 2=2010w 2=t

then x= (t 2007) y= (t 2008) z= (t 2009) w= (t 2010).

1/x+1/y+1/z+1/w=1

2007+√2008+√2009+√2010)/√t=1

t=√2007+√2008+√2009+√2010

2007x+2008y+2009z+2010w)

[√2007t)+√2008t)+√2009t)+√2010t)]

[√t(√2007+√2008+√2009+√2010)]

There is no solution to this question.

2. a^6+3a^4*b^2+3a^2*b^4+b^6=a^6+2a^3*b^3+b^6+8a^3*b^3

Vested 3a 4*b 2+3a 2*b 4=10a 3*b 3, both sides divided by a 3*b 3

3a b + 3b a = 10, i.e. b a + a b = 10 3

3.Let m=2007x 2=2008y=2009z2=2010w2, then x= (m 2007) y= (m 2008) z= (m 2009) w= (m 2010).

1/x+1/y+1/z+1/w=1

2007+√2008+√2009+√2010)/√m=1

t=√2007+√2008+√2009+√2010

2007x+2008y+2009z+2010w)

[√2007m)+√2008m)+√2009m)+√2010m)]

[√m(√2007+√2008+√2009+√2010)]

Happy New Year.

1.The answer is 1

A +1=a,a 2010+1 a 2010=[(a 2010) 2+1] (a 2010).

Let a 2010=x, the original formula = (x 2+1) x=x x=1

2.The answer is 8 3

The original title reads (a +b) (a +b) (a +b) = (a +b) (a +b )+8a b

Divide both sides by a, (1+ b a) (a+b a)(a +b ) = (1+ b a) (a +b )+8b

Divide both sides by b to get (1+ b a) (a b + b a) (a b +1) = (1+ b a) (a b +1) + 8

So, (a b + b a) = 8 3

3.The answer is 2007+ 2008+ 2009+ 2010

2007x²=2008y²=2009z²=2010w²

2007 x=√2008 y=√2009 z=√2010 w

y=√2007/√2008 x , z=√2007/√2009 x, w=√2007/√2010 x

1 x+1 y+1 z+1 w=1

Yes: 1 x + 1 (2007-2008 x) + 1 (2007-2009 x) + 1 (2007-2010 x) = 1

x=(√2007+√2008+√2009+√2010)/√2007

y=(√2007+√2008+√2009+√2010)/√2008

z=(√2007+√2008+√2009+√2010)/√2009

w=(√2007+√2008+√2009+√2010)/√2010

Root number 2007x+2008y+2009z+2010w = 2007+ 2008+ 2009+ 2010

(1) A, B and C run 800 meters along the 400-meter circular track, when A runs 1 lap, B runs 1 7 laps more than A, and C runs 1 7 laps less than A, if the speed of their respective runs is always the same, then when B reaches the end line, how many kilometers is C from the finish line?

Solution: When A runs 1 lap, B runs 8 7 laps, C runs 6 7 laps, then when B reaches the finish line, C runs x laps:

2:8/7=x:6/7

x=2*6 7*7 8= laps.

That is, C is still from the end point: 400 * (m.

2) A, B, C originally had a total deposit of 3460 yuan, because A took out 380 yuan, B deposited 720 yuan, C deposited 1 3 of his original deposit, and now the ratio of the number of deposits of the three people is 5:3:2, and the current deposits of the three people are (2000) yuan, (1200) yuan and (800) yuan.

Solution: If the current deposits of the three people are 5x, 3x, and 2x yuan respectively, then:

5x+3x+2x=3460-380+720+2x/(1+1/3)*1/3

10-1/2)x=3800

x=400, i.e. a, is 5x=5*400=2000 yuan.

b is 3x=3*400=1200 yuan.

c is 2x=400*2=800 yuan.

3) The number of soccer balls and basketballs in the school is 8:7, first buy a number of soccer balls, then the ratio of the number of soccer balls to basketball is 3:2, and buy some basketballs, then the number of soccer balls and basketballs is 7:6, it is known that there are 3 more soccer balls than basketballs, how many are there?

Solution: The school originally had 8x and 7x footballs and basketballs respectively, and bought y soccer balls and (y-3) basketballs, then: (8x+y): 7x=3:2 ......1）

8x+y):(7x+y-3)=7:6……（2）

From (1): x=2 5*y.........3）

Substitute (3) into (2) to get:

8*2/5*y+y):(7*2/5*y+y-3)=7:6

7*(7*2/5*y+y-3)=6*21/5*y

y=15x=2/5*y=2/5*15=6

That is, the original number of soccer and basketball is 8x=8*6=48, 7x=7*6=42 respectively.