Question 1

The topic is too troublesome, so I use ABCD ... instead.

The topic is A * B-C * D. Note that I changed it to the following type: A= 1+D, B does not move, C= 1+B, D does not move.

In this way, the formula becomes (1+d) * b-(1+b) * d. If you open the brackets and merge similar items, you will find that the simplified item is (B-D). Now you look at B and D together with the original data, and finally they all disappear, leaving one 1/65438+.

Question 2

If you look closely, you will find that (for convenience) Ling: 1*2*4=A,1* 3 * 9 = B.

Now the formula becomes (a+2a+3a+...+ sodium /b+2b+3b+...+nb) 2.

That is to say, both numerator and denominator become arithmetic progression. Using the summation formula of arithmetic sequence,

The numerator becomes n(n+ 1)A/2 and the denominator becomes n(n+ 1)B/2.

Divide the numerator by the denominator and then square it to get (A/B)2. Now take both A and B in and have a look. The result is 64/729.

Question 3

This is the title of a clock.

You must first know this truth: every time the minute hand turns 360 degrees, the hour hand turns 360/ 12 degrees, which is 30 degrees.

Then the minute hand turns 1 degree, and the hour hand turns 30/360 =112 degrees.

Well, now the problem becomes a catch-up problem.

At 4 o'clock, the angle between the hour hand and the minute hand is 120 degrees.

Suppose that the minute hand can only coincide with the hour hand by rotating it clockwise by x degrees. Equation:

X= 120+X/ 12, and the solution is x =1440/1,which is approximately equal to 130.9 1 degree.

Question 4

First of all, you should know that there are two answers to this question. The first is from 4: 00 to 4: 30, and the second is from 4: 30 to 5: 00.

The first case: suppose that starting from 4 o'clock, the minute hand takes X degrees and forms a 90-degree angle with the hour hand.

Column equation:

120+X/ 12-90=X, and the solution is X=360/ 1 1.

The second case: suppose that the minute hand and the hour hand form a 90-degree angle for the first time, then the minute hand can form a 90-degree angle again after passing through Y degrees.

Column equation:

Y-Y/ 12=90, and the solution is y =108/1. That is, from 4 o'clock, when the minute hand goes 360/11+108/65438.

In the clock, the minute hand goes 1 turn, which means it takes 60 minutes to go 360 degrees, so every turn of the minute hand is 60/360= 1/6 minutes.

So the time of the first case is 360/ 1 1 *1/6 = 60/11,which is 60/1minute after 4 o'clock.

In the second case, the time is 468/11*1/6 = 78/ 1, which is 78/1min after 4 o'clock.

The fifth question

One person has already answered this question, which is absolutely correct. I also use this method, which is often used in junior high school math competitions. I posted it directly:

x/ 1×2+x/2×3+......+x/ 1999×2000+x/2000×200 1 = 2000

x( 1/ 1×2+ 1/2×3+......+ 1/ 1999×2000+ 1/2000×200 1)=2000

x[( 1- 1/2)+( 1/2- 1/3)+( 1/3- 1/4)+......+( 1/ 1999- 1/2000)+( 1/2000- 1/200 1)]=2000

x[ 1- 1/2+ 1/2- 1/3+ 1/3- 1/4+......+ 1/ 1999- 1/2000+ 1/2000- 1/200 1]=2000

x( 1- 1/200 1)= 2000

X=200 1

The sixth question

It's simple. As we all know, if a positive number is added to the numerator, the fractional value will be larger and larger, and if a positive number is added to the denominator, the fractional value will be smaller and smaller, and 9/7 is obviously greater than 9/ 13. Therefore, if 0 can be added, the minimum value of a+b is that when A = 0 and B = 6, a+b=6 is the minimum. In short, to minimize a+b, we must minimize A. Since a= 1.b=67/9, it doesn't fit the question. A=2, b= ... Keep trying. Let me help you calculate when a=9 and b= 19. So a.

Well, I worked hard to calculate, write and get the answer for you. It took me several hours. I hope it helps you. If you think it's ok, can you send me more? This is hard work. Thank you.