Olympiad questions in the first year of junior high school Students who are good at Olympiad mathema

Landlord, this is quite simple, you see, assuming that each cow eats one serving of grass a day, 24 heads for 6 days is 24 6 = 144

21 first 8 days is 21 8 = 168 parts, so it can be calculated that a total of 168-144 = 24 parts grow in 2 days, and on average, 12 parts of grass grow every day, so that 144-12 6 = 72 parts (or 168-12 8) is used to determine the number of original grasses.

1. Graz 16 cows, let 12 go to eat the new grass that grows every day, and 4 eat the 72 servings, so 72 4 = 18 days.

2. Let never finish eating, only 12 servings can grow every day, and the remaining grass will always be eaten one day no matter how many cows eat, so at most the cows can only eat the newly grown grass, and never finish eating, then up to 12 heads.

1) Suppose 1 head eats 1 unit a day, then.

16 cows minus 12 heads, leaving 4 heads to eat the grass of the original pasture:

72 (16 12) 72 4 18 (days) So raise 16 cows, and it takes 18 days to eat up all the grass on the pasture.

2) Grass eaten by 24 cows for 6 days: 24 6 14421 cattle eaten for 8 days: 21 8 168

1 day of new grass: (168 144) (8 6) 12 The original grass on the pasture is: 24 6 12 6 72 The new grass grows every day is only enough for 12 cows.

Therefore, in order to never finish this grass, you can only put a maximum of 12 cows to eat this grass.

Set the original grass x in the pasture, the growth rate v, and each cow eats grass p every day; Then:

x+v*6=24*6*p ①

x+v*8=21*8*p ②

1) Set n days to eat.

x+v*n=16*n*p ③

Subtracting from the equation yields: v=12p

Dividing by the equation yields x=6v=72p

Bringing x and v into the formula: 72p+12p*n=16*n*p, about p on both sides, can be obtained n=18 days.

2) In the same way, let m cows be eaten in n days.

x+v*n=m*n*p ③

n=72/(m-12)

It can be seen that when m<=12, n is negative or infinite.

Therefore, up to 12 cows can be placed.

168-144) (8-6) = 12 (the amount of grass growing per day) 144-12 * 6 = 72 (original grass).

1) 72 (16-12) = 18 days.

2) 12 cows.

Since the sum of the three numbers is known, if you want to add the smallest number of "Hua Cup Preliminary Round", as long as the other two add the largest.

And because the digits of the above two additions are represented by different Chinese characters, repeated numbers cannot be used, as long as 9 is placed on the hundred place, 8 and 7 are placed on the ten place, and 6 and 5 are placed on the single place, so that the sum is the largest.

And for: 986 + 75 (of course, such as 976 + 85 can also be) = 1061 so the minimum value of the "Hua Cup Preliminary Round" is 2011-1061 = 950 One less thing, sorry.

Since the minimum value of Huabei can be 950 and four digits, it must be represented by the smallest four digits (numbers are not repeated), which is 1023

The hundred digit of sum is 0, so there must be a carry, so "Hua" = 1

It is known that a+(bc-a) (a+b +c)=b+(ac-b) (b +a +c).

Simplification yields (a-b)(a+b+c)=(a-b)(a+b+c)a,b are unequal, so a+b +c = a+b+c.

Thus, A+(BC-A) (A+B +C)=A+(BC-A) (A+B+C)=(Ab+BC+Ca) (A+B+C), obviously the values of the exchange A and C eras remain unchanged.

If a+b+c=1, then (a+b+c) =1, a +b +c +2(ab+bc+ca) = 1

So ab+bc+ca=0. Therefore, the value of the original algebraic formula is 0 (fixed value).

I wonder if I can solve a binary equation in the first year of junior high school?

1. Set bag A as x and bag B as y

Then 1 2*x=y-12

1/2(y-12)=1/3*y

So y=36, x=48

a total of 84kg

2. Set A X, B Y

1)x+y=98

When A is 1 2y, B is exactly x

then the time is (y-x) years from now.

There is 1 2y+(y-x)=x

i.e. (2) 3 2y=2x

x=42,y=56

3. Set up an x-disk under this day.

In the morning, in the afternoon x-(

So there is 2+x=10

4. (1), A * 3 + B * 7 + C = 315

2), A * 4 + B * 10 + C = 420

So A + B * 3 = 105

3) 3 * A + 9 * B = 315

2)-(3) Deaf.

A + B + C = 105 yuan.

5. Set the length x, eat y, it turns out that the grass has a

24*6*y=6*x+a

21*8*y=8*x+a

x=12ya=144y

16y*t=144y+x*t

t=36 days.

2) Explain that if you want to never finish eating, then the grass that grows every day is at most equal to the grass that is eaten, that is, the grass that grows is just eaten, and you can put up to 12 cows from the above.

Set up n cows.

a+x*t-n*y*t>0

144y>(n*t-12t)*y

144>(n-12)t

Because t-> infinity, n takes a maximum of 12, that is, a maximum of 12 cows.

PS: The above is just the most common solution, which is relatively easy to understand.

Since it's an Olympiad problem, then I don't need equations, I remember that I was not allowed to use equations 1, B took the remaining half for the second time, and there was 1 3 left, so the first time I took it, B still had 2 3, so B bag had a total of 36kg

B takes 12kg and weighs half as A, so A has 48kg, i.e.: 12 (1-1 3-1 3 (1 2)*(1-1 2)) = 36kg

2 When A's age is reduced to half of B's age, B is A's current age, and they have reduced (or increased) B's age, respectively: 2- The age difference has decreased B's age-age difference*2 so many years, and the age and value at this time are A's age + half of B's age.

Therefore, age A + age B = 98 years old times B age + A age - 2 times the age difference is still 98, so B's age is 4 times the age difference, A's age is 3 times the age difference and times the problem, so the age difference is 98 7 = 14 years A is 42 years old B is 56 years old 3 4 5 It seems that only the equation can be used.

You can buy a book of special exercises for top students, and the topics in it are quite good.

The topic is not difficult, you should see that the relationship between the former and the latter is consistent, so the key is that we use this relationship, so you see the adjacent two as a whole, then the internal structure of each whole has consistency, so you can use multiple groups to study as a group, as long as the other hail groups determined by the common relationship will also have this relationship, so you see ax (x no matter how much) as an unknown quantity x or y, then the consistency relationship of each group can be described like this, (x, y) (|y|=|x+2|Then we see that the question of the wide and disadvantageous sail problem requires the minimum value of multiple x+y, so we use the relationship between x and y to solve it, and use the absolute value inequality, which is: x+y>=|x|-|y|=|x|-|x+2|> = 2 (average amount -2 2 = -1), so a1 + a2 +. a2005=（a2+a3）+（a4+a5)..

a2004+a2005)>=1*2004=-2004, and then we can get it by looking at the equal sign, such as a1, a3, a5....a2005=0，a2，a4，a6.。。a2004=-2, i.e., it is obtainable.