The master solves the Olympiad urgent .

If A, B and C each win 1 game, then D wins 3 games, obviously D has lost to A, so A, B and C each win 2 games, and D wins 0

A win at least 1 game.

If A wins 3 games, B wins B. B C is unlikely to win three games.

If A wins 1 game, then B loses to B C. If B and C only win 1 game, then B and C only win A, and no one can win the game B vs C, which is contradictory.

So if A, B and C win 2 games, and a total of 6 games are played. Ding won 0 games.

A: First of all, A can't win 3 games. Because if A wins 3 games (B, C, D), then B can only win 2 games at most, which is not the same as the number of games won by A.

If A wins 2 games (let's say B wins B and D and loses to C), then B should win the same number of games as A, which is also 2 games. Since B loses to A, he can only win against C and D. C, also won 2 games, he won A, lost to B, but he won D.

So, Ding won 0 games here.

If A only wins 1 game (wins D), then A loses to B and C, and B and C can only win 1 game, i.e. they win A. If B and C draw and both lose to D, then in line with the title, D wins 2 games here (wins B and C). If B and C are divided into winners and losers, it does not fit the topic.

So, the possible outcomes are:

A wins B; A loses C; Jia Shengding;

B wins C; Yi Shengding; Cingsheng Ding.

So Ding won 0 games.

Answer: 2500 RMB.

A: You can take out the yuan.

h does not carry. B+C can only go up to 1

G+F can only go up to 1

And the last sum of thousands is not d, i.e., d+1=a, and g+f>10

d≠1，a≠1,g≠1,f≠1

cei*d=dfb

c=1, otherwise the hundredth cannot be d

e*d<10, otherwise to be carried forward, the number of hundreds, greater than one and the product less than 10 of the two numbers may be 2*3, 2*4

d can only be 2 or 3 or 4

c+b=f or c+b=10+f

b+1=f or b=9+f, i.e. b=9, f=0

When b=9, f=0, CEI*d=d09

If the single digit is the product of 9, the multiplier and the multiplied unit can only be 1*9, 3*3, 7*7

c = 11 * 9 exclusions.

The other letters are different.

3*3 and 7*7 exclusions.

b+1=f<10

f+g=10+d

Continue looking at CEI*d=dfb

Suppose i*d is not carryed, i.e. i*d=b

According to 2), e*d is not carryed, i.e. e*d=f

According to 3) b+1=f, i.e. (e-i)*d=f-b=1, it obviously does not match the title.

i.e. i*d carrying.

Suppose i*d carries x

That is, i*d=x*10+b, and e*d+x=f

When e*d=8, x=1, f=9

i.e. e=2, d=4 or e=4, d=2

When the lack of rot excitation e=2, d=4, a=5

then i is even, otherwise h=5=a

When i=6, CEI=126, DA=45, 126*5=630, i.e., G=6, C=3, contradiction, exclusion.

When i=8, CEI=128,128*5=640, i.e., g=6, c=4, contradiction, exclusion.

When e=4, d=2, a=3

cei=14i，da=23

i=5, h=5, exclude.

i=6, 146*23=3358 excludes.

i=7, h=1, exclude.

i=8, h=4, exclude.

i=9, 149*23=3427, d=4, excluded.

e*d≠8 When e*d=6, that is, e=2, d=3 or e=3, d=2, and d+1=a

d≠2, otherwise e=a=3

d=3,e=2

cei*d=12i*3=3fb

Stockings have the numbers 1 4

i 5 When i = 5, b = 5, exclude.

When i=6, b=8, f=7, contradicts b+1=f, excludes.

When i=7, b=1, exclude.

When i=8, b=4, exclude.

When i=9, b=7, f=8, yes.

At this time, CEI=129, DA=34, GCH=516, DFB=387, ADFH=4386, i.e.:

a=4, b=7, c=1, d=3, e=2, f=8, g=5, h=6, and i=9

Exactly in line with the topic, that is, what you want.

Measure the depth of the lake with a rope and fold twice into the bottom of the lake The remaining rope on the water is 1.5 meters, if you fold 3 times into the bottom of the lake, the length of the rope is less than 5 meters How deep is the lake?

Analysis]: What does not change in this problem is the depth of the well, and the equation can be obtained by expressing the depth of the well with an algebraic formula The equivalence relationship in this problem is:

Fold 2 times into the bottom of the lake, and the remaining rope on the water is 1.5 meters;

Fold 3 times into the bottom of the lake, the length of the rope is less than 5 meters.

Answer]: Solution: The length of the rope is x meters, and the depth of the well is y meters.

According to the fold 2 times into the bottom of the lake, the remaining rope on the water is 1.5 meters: then y=x 2 fold 3 into the bottom of the lake, the rope length is less than 5 meters, then y=x 3+5

Measure the depth of the lake with a rope, fold it twice into the bottom of the lake, and the remaining rope on the water is one and a half meters, and if it is folded into the bottom of the lake three times, the length of the two ropes is less than 5 meters.

Q: How deep is the lake?

Solution: If the rope is often x meters, and the lake is y meters deep, then there is an equation:

y=x/3-3/2...1)

y=x/4+5...2)

From (1) (2) x 3-3 2 = x 4 + 5, that is, there are 4x-18 = 3x + 60, so x = 78 meters; y=78 3-3 2=m.

That is, the rope is 78 meters long and the lake is deep.

Solution: Let the length of the rope be xm

x solves: x=39

Water depth: 39

A: The depth of the lake is 18 meters.

I'm sorry I didn't read the question just now.

Solve the water depth xm

2（x+2x+3=3x-15

3x-2x=15+3 Your classmate made a mistake in this step x=18

If there are still doubts.

Please type out the question exactly as it is.

.Let the length of the rope be l and the depth of the lake be h, then there is: h=l ; h=l 5+5 is solved: l= h=

The depth of the lake is meters. 2.Let the length of the rope be x and the circumference of the trunk be y, then there is: y=x 3-3;y=x/4+7

The solution is: x=120 dm y=37 dm, the length of the rope and the circumference of the trunk are 120 and 37 dm, respectively.

49*3=147 147-49-49=49 49 6=8 So a minimum of 9 votes is required to be elected.

Solution: From the question, analyze the distance between the two meeting places, which is the distance traveled by Xiao Zhang's 1 hour and 15 minutes; (1 hour 15 minutes = hours) to find Xiao Zhang's walking speed:

Explain that Xiao Zhang has walked a total of 1 5+

Xiao Li rides a bicycle at a speed of:

The distance that Xiao Li traveled by bicycle:

The circumference of the circle is:

Set the perimeter to eliminate orange x

Speed and =x 1=x

Speed difference = x

Cyclist Oak Speed Pick Potato Ball:

Walking speed: x=50km

[n^3-(n-1)^3]-[n-1)^3-(n-2)^3]}-n^3-2(n-1)^3+(n-2)^3-[(n-1)^3-2(n-2)^3+(n-3)^3]

n 3-3 (n-1) 3+3 (n-2) 3-(n-3) 3 (careful) n 3-3n 3+9n 2-9n +3+3n 3-18n 2+36n-24-n 3+9n 2-27n+27

The original proposition has been proven! Scrambled ridges.

The slow dust infiltration you need to understand is (x y) 3=x 3+3x 2y+3xy 2+y 3.

If you don't have a self-study of junior high school content, you really can't do it, and it's hard to explain Li Zheng clearly to you! It is recommended that you first teach yourself junior high school algebra. Now the textbooks are different in different places, so I can't say which chapter I specifically studied by myself, you can teach yourself the addition, subtraction, multiplication and division of algebraic formulas.

Because the grass on the three meadows is the same thick and grows just as fast, the inscription shows that each hectare of land can feed 6 cows for 6 weeks, and 4.5 cows for 12 weeks, and only need to calculate that each hectare of land can feed 5 cows for x weeks.

Let each cow eat a per hectare of soil per week, and grow b grass per week, initially c has 6*6a=c+6b 1

2 is subtracted from 2 to 1 to get b = 3a

Substituting 1 gives c=18a

Then 5xa=c+xb gives x=9

So it can feed 50 cows for 9 weeks.

Answer: Compare on the same area of grass.

The first meadow is 4 hectares for 24 cows for 6 weeks, then the first meadow for 1 hectare is 24 4 = 6 cows for 6 weeks.

The second plot of grassland is 8 hectares and can feed 36 cattle for 12 weeks. Then the second meadow 1 hectare can feed 36 8 = cattle for 12 weeks.

Solution: Suppose the amount of grass eaten by each cow per week is 1.

6x6 = 36 servings.

Portion. Grass growth per hectare per week: (54-36) (12-6) = 3 portions.

Initially there is grass per hectare: 36-3x6 = 18 parts.

10 hectares of grassland grows 10*3=30 per week

18 servings * 10 hectares (50 servings - 30 servings) = 9 weeks.

Let's say each cow eats 1 serving of grass per week.

The first plot, 24 cows, 6 weeks, a total of grazing: 24 6 = 144 servings.

The area of the second plot is 8 4 = 2 times that of the first plot.

Feeds 24 2 = 48 cows for 6 weeks.

Grass eaten in total: 48 6 = 288 servings.

Now it is 36 cows for 12 weeks, a total of grazing: 36 12 = 432 servings.

Difference: 432-288 = 144 parts.

These 144 are the grass that grows in the second plot of land in 12-6=6 weeks.

Average weekly long grass: 144 6 = 22 servings.

It turned out that there was grass: 288-22 6 = 144 parts.

The area of the third plot is that of the second: 10 8 = 5 4 times.

It turned out that there was grass: 144 5 4 = 180 parts.

Long grass per week: 22 5 4 = servings.

50 cows come to eat, in addition to eating the newly grown portions every week.

And eat the original: portions.

Feeds 50 cows: 180 weeks.

Let x = 0, find a = 10

Let x = 1, and just get a = 10, find b =

Sou which state pants to a trace Chang - b Li Jian =