Primary School Olympiad Short Calculation Questions, Primary School Olympiad Mathematics Questions,

7.(of the loop).

8.(2 and 1 2003 9 and 5 8 + 7 2002 and 2003 and 1 4.)

9.2002 2002 2002 2003 1 20047(of the loop).

8.(2 and 1 2003 9 and 5 8 + 7 2002 and 2003 and 1 4.)

9.2002 2002 and 2002 2003 1 2004 recurring equation calculation).

Equal to 12

7.(of the loop).

8.(2 and 1 2003 9 and 5 8 + 7 2002 and 2003 and 1 4.)

9.2002 2002 2002 2003 1 20047(of the loop).

8.(2 and 1 2003 9 and 5 8 + 7 2002 and 2003 and 1 4.)

9.2002 2002 2002 2003 1 2004

You can't use your brain if you don't know how to know, you can't ask for help from others, and don't develop the habit of dependence.

Let's start with a more common topic, take the topic I talked about on a small piece of paper as an example:

First of all, the reason why we can cross out a 0 of the dividend and the divisor at the same time is because we follow the rule that "the dividend and the number of jujube yards are reduced by the same multiple at the same time, and the quotient does not change". So, if you remove a 0 at the same time, it doesn't affect the result, so the original 65100 210 now becomes 6510 21, but the result is the same. Calculate 6510 21 and the final result is 310, so 65100 210 310.

Here's an example of a mistake: pay attention to the 0 at the end of the quotient, don't write it, because although the single digit still exists, there is no longer a number on the single digit.

Let's look at the second example.

3620÷50＝72...20, this is what we did with the previous method, there are no problems, the final result is 72 out of 20, but, looking further down, the new easy way.

Here's the new easy way: the final remainder is 2, is there something wrong with our calculations? Or is our new approach wrong?

In fact, it's not, how to understand it, first of all, the divisor is reduced by the dividend by the same multiple, the quotient is unchanged, and Yan Xiao is to say, in the vertical type, the quotient is a few, we can copy it directly, but the remainder is different, as I said above, although the 0 in the dividend is crossed out, but the digit is still there, our remainder 2, on the ten, represents 2 10, so it is 20. Summed up in one sentence:

Use a simple way to do pen calculations, the quotient is a few, directly copied in the horizontal back, but the remainder must be careful to see which digit it is in, like this topic, 2 in the ten digit, it represents 20, so the remainder 20.

Why are they all the same kind of questions...

The solution is the same. For example: (1) primitive = [(1 2)-(1 4)+(1 4)-(1 6)+(1 6)-(1 8)+(1 8)+1/48)-(1/50)]/2=(1/2-1/50)/2=6/25

2) Original formula = 1 1-1 4+1 4-1 7+1 7-1 10+1 10-1 13+1 13-1 16=1-1 16=15 16

Similar (3) 16 99 (4) 1 20 (5) 3 200

Let a = (1/2 + 2/3 + 3/4 + 4/4 + 4/5 + 5/6 + 6/7), then the original formula becomes:

a²＋½a－(1+a)×(a-½)

Got: a + a -a + a - a - a +

Use the commutation method. Let a=(1/2 + 2/3 + 3/4 + 4/4 + 4/5 + 5/6 + 6/7) and substitute it for the original formula.

a² +1/2a - 1+a) x (a-1/2) =a² +1/2a - a + 1/2 - a² +1/2a = 1/2

I don't think I taught 2 power in elementary school.

Let a = (1/2 + 2/3 + 3/4 + 4/4 + 4/5 + 5/6 + 6/7), then the original formula becomes:

a a (1+a) (a-) Description: The term related to a is later offset).

End Result

These are different forms of practice questions of the same type.

Let me briefly talk about the method!

This calculation of splits and splits.

Divide each term into two fractional difference forms, and then sum them, so that the subtraction of the first term and the subtraction of the second term can be canceled out, so that at the end you can eliminate the infinite number of terms in between, leaving the head and tail. And then calculate.

Analysis: 1 2*4 can be split into 1 2 -1 4 The result is (4-2) (2*4) is 2 times worse than 1 2*4. Finally, divide by 2

In this way, each term is divided by 2 to extract the common factor, 1 2

Analysis: The numerator is directly the difference between the two factors of the denominator, and it is enough to split it directly. 3/1*4=1-1/4

This is similar to the first question.

This is similar to the first question, but there is a difference of 8-5=3 times.

This is also similar to the first question, with a difference of 15-10=5.

Why are they all the same kind of questions...

The solution is the same. For example: (1) primitive = [(1 2)-(1 4)+(1 4)-(1 6)+(1 6)-(1 8)+(1 8)+1/48)-(1/50)]/2=(1/2-1/50)/2=6/25

2) Original formula = 1 1-1 4+1 4-1 7+1 7-1 10+1 10-1 13+1 13-1 16=1-1 16=15 16

Similar (3) 16 99 (4) 1 20 (5) 3 200

The last simplification of the lz man was wrong, but why is your answer correct? I can't figure it out.

The correct solution should be:

Solution: It can be known from the meaning of the title.

Every project is.

1-3/(n-1)(n+1)

(n-1)(n+1)-3]/[(n-1)(n+1)]=(n*n-4)/(n-1)(n+1)

(n-2)(n+2)] [(n-1)(n+1)] The original formula can be obtained in the above way.

Original = (1*5) (2*4) *2*6) (3*5)*(3*7) (4*6).95*99）/（96*98）*（96*100）/（97*99）

Then the n-1 of each term and the n-2 of the last digit are approximated, then the n+2 of each term and the n+1 of the last digit are approximated, and finally the remaining part is.

1-3/(n-1)(n+1)

(n-1)(n+1)-3]/[(n-1)(n+1)]=(n*n-4)/(n-1)(n+1)

(n-2)(n+2)] [(n-1)(n+1)] The original formula can be obtained in the above way.

Original = (1*5) (2*4) *2*6) (3*5)*(3*7) (4*6).95*99）/（96*98）*（96*100）/（97*99）

Then the n-1 of each term and the n-2 of the last digit are approximated, then the n+2 of each term and the n+1 of the last digit are approximated, and finally the remaining part is.