Mathematics problems in the first year of junior high school!!

1) The absolute value of 1/102-101 plus the absolute value of 1/103-102 is also reduced by 1/103 minus 101/101) absolute value.

2）.Observe the following equation: 1 2 1 integrals + 2 3 1 integrals + 3 4 1 integral = 1-2 + 2 - 1 3 + 1 3 - 1 4 = 1 - 4 = 1 - 4 / 1 = 3 quarters.

Guess and write: n(n+1) 1/1 = 1 n-1 (n+1) (fill in).

3）.One of 1 2 + one of 2 3. Omitted) + one of the points of 2011 2012.

4）.One of 1 2 + one of 2 3. Omit) + n (n + 1).

1-1/（n+1)

n/(n+1)

5）.One of 2 4 points + one of 4 6 points. 2010 One of the points of 2012.

Question 1 |1/102-1/101|+|1/103-1/102|+|1/103-1/101|

Question 2: 1 n(n+1)=1 n-1 n+1

Question 3: 1 1 2 + 1 2 3 + ......1/2011 2012

Solution: Suppose x kilograms of candy A and candy Y B are required.

Then I solved the equation in the graph, and I calculated that it was x=50 and y=150

A: ...

If A is x kg, then B is 200-x kg, then x =

50 kg. 200-x = 150 kg.

Let the first type of candy be x kilograms.

x=50A50 B150

1. If the train speed is x km h, then (x+

If the solution is x=, then the train length is (

2. The walking time of teachers and students in the school is x minutes, and if the car is moving, its speed is 6 times the walking speed, and the distance is twice the round-trip, and the time is x 3 minutes. Car repair takes t minutes.

According to the title: x-x 3+10+t=30

I can only come up with this one equation, and my ability is not enough.

1: Set the train x meters.

The speed of the train is (x-1) 15=(x+1) 17x=16 meters.

2: Half an hour.

Teachers and students get on the bus on the way to the county seat, so no matter how many miles the teachers and students walk, the speed of the car remains the same, and the time when they are late is the time for car repairs.

1: Set the train x meters.

The speed of the train is (x-1) 15=(x+1) 17x=16 meters.

2 solution: If the distance between the school and the county seat is S kilometers, the car failure time is a minute, the walking time of teachers and students is t minutes, and the walking speed is X kilometers minutes, then the speed of the car is 5x kilometers minutes, 10+t+s-xt5x=

s5x+30，a+2(s-xt)5x=

2s5x+30, the solution: t=25, a=40, after testing: t=25, a=40 is the solution of the original equation

A: It took a total of 40 minutes for the car to troubleshoot on the way

After separation, A and B walked in opposite directions along the railroad tracks. At this time, a train is heading towards A at a constant speed, and the train passes by A's side in 15 seconds; Then I drove next to B and it took 17 seconds. How long is the train when it is known that both people are walking at kilometers at a speed of one kilometer?

Analysis: Towards A, the train took 15 seconds to drive next to A, here is the encounter problem, and the encounter distance is the length of the train.

It took 17 seconds for B to drive beside him, and because A and B were separated and went in the opposite direction along the tracks, then the train was chasing B at this time, and the distance to catch up was the length of the train, and the equation was listed according to the analysis:

Conversion unit: human speed: meters and seconds.

Solution: Let the speed of the train be x

15（x+1）=17（x-1)

x=16s=(16+1)*15=255m.

Due to various reasons, Ah Q is always unlucky in horse trafficking production. Once, Ah Q bought a bad horse for 780 yuan, and after feeding it for a period of time, he sold it for 1800 yuan. The loss is exactly half of the purchase price of the bad horse plus a quarter of the feed price.

Q: How much money did Ah Q lose?

According to the title: 860 yuan.

Set the feed money x yuan.

then there is x (1800-780) 780 2+x 4 to get x 1880

Solve the loss of money x-(1800-780) 860 thank you for adopting.

Set the feed money is $x.

780+x=1800+780÷2+¼x

x=1880

Ah Q lost 860 yuan.

Set feed x yuan.

1800-780-x=390+1/4x

4/5x = 630

x=504, so 780+504=1284 (equivalent to the total purchase price), 1800-1284=526 (equivalent to profit).

780-526 = 274 (equivalent to lost money).

That is: Ah Q lost 274 yuan.

Assuming that the feed money is x yuan, the relationship between the money after Ah Q sells the horse is as follows:

780+x money spent on horses and feed for Ah Q;

1800 is the money for the sale of horses, then the relationship is:

780 + x - 1800 = 780 * (1 2) + (1 4) * x solution to get x = 1880

So the loss is 1880 * (1 4) + 780 * (1 2) = 860

Set the feed money x yuan.

780+x=1800+780/2+x/4

x=1880

So, 780 2+x 4=860

Therefore, Ah Q lost 860 yuan.

According to the title:

Compensation of 860 yuan.

Set the feed money x yuan.

then there is x (1800-780) 780 2+x 4 to get x 1880

and lost money x-(1800-780) 860

Set feed x yuan.

Lost 780+x-1800

Again, as follows. Lost 780*

780+x-1800＝780*

x＝1880

Lost 1880 + 780-1800 860

From the position on the problem and the number line, we can find b=-1 and c=2, so it is |2a+3|-|3-2a|+|1|+2=|-1|+2=1+2=3

Big brother: I can't see the picture.

Uh·· I can't see the picture, how can I do it? You can I send me a picture.

Can you explain the order of the points from left to right?

As can be seen from the figure.

b＜0，c＞1

So 6|b|=-6b=6, so b=-1

3|c|=3c=6, so c=2

then |2a-3b|-|3b-2a|+|b|+[c）]=|2a+3|-|3-2a|+1+2

2a+3|-|2a+3|+1+2=3

Answer: Let the original price of commodity A be x yuan, and the original price of commodity B is y yuan

x+y=500

Solution: x= 320 y=180

A: The original prices of the two products are: 320 yuan, 180 yuan.

70 : 90 A: 500*7 7+9 B: 500*9 16=7:9 =500*7 16 = yuan).

Yuan) I use the ratio to calculate, I don't know if it's right (I'm only in the 6th grade, it's best not to believe it.)

Solution: Set the original price of product A to be $ x, and the original price of product B to be $ y.

Let x+y=500

70%x+90%y=386 ②

Result: y=500-x

Substituting gets: x=320

Substituting x=320 yields: y=180

x=320y=180

The result should be like this, if you don't believe it, go in and check it yourself.

If the price of the goods is x yuan, then 70 x + 90 (500-x) = 386, and the solution is x = 128

Set the goods after the seven-fold discount is X yuan:

then 70 x+90 (500-x)=386

Solution x=128