Mathematical equation of first degree with one variable A. There are two routes between two places B. one person rides a bicycle along Route 1 to B at the speed of 9km / h, and then returns to place a from place B along Route 2 at the speed of 8km / h. It is known that route 2 is 2km less than Route 1, and it takes one eighth of an hour less. How long is Route 1? There are several black and white balls, and the same color balls have the same mass, so do the weights. For the first time, one black ball plus two white balls is equal to one weight, and for the second time, three black balls plus one white ball is equal to two weights. How many white balls can be balanced with one black ball?

Mathematical equation of first degree with one variable A. There are two routes between two places B. one person rides a bicycle along Route 1 to B at the speed of 9km / h, and then returns to place a from place B along Route 2 at the speed of 8km / h. It is known that route 2 is 2km less than Route 1, and it takes one eighth of an hour less. How long is Route 1? There are several black and white balls, and the same color balls have the same mass, so do the weights. For the first time, one black ball plus two white balls is equal to one weight, and for the second time, three black balls plus one white ball is equal to two weights. How many white balls can be balanced with one black ball?

(1) Let the length of route one be x, then the length of route two be X-2
According to the meaning of the title:
X/9=(X-2)/8+(1/8)
The solution is: x = 9
So the length of route one is 9km
(2) The weights have the same mass, so according to the meaning of the title, the mass of the second time is twice that of the first time
So: let y white balls and a black ball balance
(Y+2)*2=3Y+1
The solution is y = 3

(Note: the following problems are solved by linear equation of one variable)
A. The distance between B and a is 176km, one of which is blocked by a landslide. Project a and B are instructed to start from a and B at 8 am and rush to the landslide site to clear the road. At 10 am, team a arrives and starts working immediately. Half an hour later, team B arrives and quickly goes into "battle" to work with team A. at this time, the additional team completes 1 / 24 of the project quantity
1. If the damaged road is 1km long, the speed of team a is 3 / 2 times that of team B, more than 5km
2. Assuming that the two teams can complete the task of dredging the road at 4 p.m. and "meet" successfully, how long will it take to complete the task if only team B dredges the road?

1. If B's speed is x, then a's speed is 3 / 2 * x + 5
That is, 2 * (3 / 2 * x + 5) + 2.5x + 1 = 176
3X+10+2.5X+1=176
5.5X=160
X=320/11
So the speed of a is (2 * 320 / 11) + 5 = 655 / 11
2. Let B work at a speed of M. from the question, we can see that a works for 6 hours, B works for 5.5 hours, and B works for (1 / 24 * 2) = 1 / 12 of the total project per hour
5.5m+(1/12)*6=1
The solution is m = 1 / 11
That is, it takes 11 hours for B to complete it alone

It can't be solved by linear equation with one variable
To build a canal, the engineering team has built 7 / 8 km, which is 1 / 4 km less than 3 / 4 of the total length. How many km is the total length of the canal? Please give the steps and ideas to solve the problem. I know the answer,

(7 / 8 + 1 / 4) / 3 / 4
It has been repaired by 7 / 8 km, and it is still 1 / 4 km short, which is 3 / 4 of the total length,
Then 7 / 8 + 1 / 4 is 3 / 4 of the total length
If you know the part and its percentage, divide the part by the percentage to get the whole

As shown in the figure, triangle ABC and triangle Dec are isosceles right triangles, a and E are right angles and equal points, and the shaded part is a square. If the area of triangle Dec is 24 square meters, then the area of triangle ABC is______ Square meters

Because the area of Figure 1 = the area of Figure 2, the area of Figure 1 + the area of Figure 2 = the area of Figure 3 = the area of Figure 4, the area of Figure 3 + the area of Figure 4 = the area of shadow part, the area of triangle Dec = the area of figure 1 + the area of Figure 2 + the area of Figure 4 + the area of shadow part =

lne^(-k)=ln（37/47）
-k=ln（37/47）
Why can I jump to the next step

lne^(-k)=loge[e^(-k）]=-kloge(e)=-k=ln（37/47）

Calculate 3.6 times 42.3 times 3.75 minus 12.5 times 0.432 times 28
Simple calculation!

Original formula = 108 × 4.23 × 1.25-1.25 × 4.32 × 28
=4.23×1.25×（108-28）
=4.23×（1.25×80）
=4.23×100
=423

The natural number (except 0) can be divided into two parts according to the number of factors

1. Prime number and composite number

Given that f (x) = (1-3 ^ 2) / (1 + 3 ^ 2), if the equation f (x) - 8 / 5sin θ = 0 has a constant real number solution on X which belongs to [- 2,2], the value range of θ is obtained

It should be f (x) = (1-3 ^ x) / (1 + 3 ^ x),
F (x) = [2 - (1 + 3 ^ x)] / (1 + 3 ^ x) = 2 / (1 + 3 ^ x) - 1, is a decreasing function,
Moreover, f (x) = (8 / 5) sin θ has a solution on X ∈ [- 2,2],
That is sin θ = (5 / 8) f (x), X ∈ [- 2,2]
So (5 / 8) f (2) ≤ sin θ≤ (5 / 8) f (- 2)
That is - 1 / 2 ≤ sin θ ≤ 1 / 2
2K π - π / 6 ≤ θ ≤ 2K π + π / 6 or 2K π + 5 π / 6 ≤ θ ≤ 2K π + 7 π / 6

5 / 8 to 0.25 is the simplest integer ratio ()

Good boy
The ratio of 5:8 to 0.25 is (5:2) and the ratio is (2.5)

Take any 51 of the 100 natural numbers from 1 to 100. It is proved that there are two numbers in the number taken, one of which is a multiple of the other

Because there are only 50 odd numbers between 1 and 100, at least one of the 51 numbers taken out is even. Because each number can be written as the power of 2 multiplied by the odd number, and there are at most 50 odd numbers, after all 51 numbers are written as the power of 2 multiplied by the odd number, there must be two numbers with the same odd factor, The difference only lies in the power of 2. So one of the two numbers with the same odd factor must be a multiple of the other