How to do this Olympiad problem? There are steps

Because (a1+b1)(a1+b2)=(a2+b1)(a2+b2) is obtained by adding (a1+b1) (a2+b1) to both sides of this equation.

a1+b1)(a1+b2)+(a1+b1)(a2+b1)=a2+b1)(a2+b2)+(a1+b1)(a2+b1) The common factor is extracted on both sides of the above equation and obtained.

a1+b1）（a1+b1+a2+b2）=（a2+b1)（a1+b1+a2+b2）

But the title says A1≠A2, that is.

a1+b1）≠（a2+b1)

So there is only one possibility, a1+b1+a2+b2=0, that is, (a1+b1)(a1+b2)+(a1+b1)(a2+b1)=(a1+b1)(a1+b1+a2+b2)=0

Because (a1+b1)(a1+b2)=1, (a1+b1)(a2+b1)=-1

It can also be proved that (a1+b2)(a2+b2)=-1

Take 3 equations to get 2 equations:

b1+b2）a1+b1b2+a1^2=1

b1+b2）a2+b1b2+a2^2=1

Find (b1+b2) and b1*b2 (only a1≠a2 have a unique solution, otherwise (b1+b2) and b1*b2 2 variables, one equation, cannot find a constant solution).

Considering b1 and b2 as solutions of the 2nd order equation: x 2 + (a1 + a2) x + a1 * a2 + 1 = 0, we get the conclusion: (a1+b1)(a2+b1)=(a1+b2)(a2+b2)=-1

a1+b1)(a1+b2)=a1^2+(b1+b2)*a1+b1b2=1 (1);(a2+b1)(a2+b2)=a2^2+(b1+b2)*a2+b1b2=1(2).

Using equation (1)-(2), we can obtain: (a1-a2)(a1+a2)+(b1+b2)(a1-a2)=(a1-a2)(a1+a2+b1+b2)=0, because a1 is not equal to a2, so a1+a2+b1+b2=0So a1+b1=-(a2+b2).

So the sought =-(a2+b2)(a2+b1)=-1

Xiao Wang departed at 8 o'clock and arrived at 9 o'clock, which took 1 hour.

Xiao Li departed at 8:15 and arrived at 9 o'clock, which took 45 minutes, so the speed ratio of Xiao Wang and Xiao Li was 45:60=3:4

Let Xiao Wang's speed be 3a, Xiao Li's speed is 4a, and Xiao Zhang's speed before catching up with the cyclist is 4a*, and then the speed is 5a*(1+20%)=6a

Let the cyclist have traveled a distance b before Xiao Wang's departure, and the speed is c, so there is 15*3a=b+15c

15*4a=b+30c

The solution is a=c, so b=30a, so the cyclist starts at 7:30.

Set off at 8 o'clock x and chase the cyclist at y o'clock.

30+y)a=5a（y-x）

5a(y-x)+(60-y)*6a=60*3a(30+y)a+(60-y)*6a=60*3a30a+ay+360a-6ay=180a

5ay=210a

y=42x=

So Xiao Zhang set off at 8:27:36.

Xiao Li: v Xiao Wang:

Xiao Zhang: , cyclist: (1 3-1 4) * (3 4) v 1 4) =

When Zhang catches up with the cyclist, the distance from place B should be twice the distance between Xiao Wang and place B at this time.

When Zhang catches up with the cyclist, the distance from place B should be twice the distance between Xiao Wang and place B at this time.

Xiao Wang caught up with the cyclist at 8:15, Xiao Wang's speed was 3 times that of the cyclist, Xiao Wang walked the back 45 minutes of the 3 5 hours, the cyclist walked the back of the 1 5, this is Xiao Zhang caught up with the cyclist when the distance from B should be 2 times the distance between Xiao Wang and B at this time.

At this time, it is 8:15 and 45 * (3 5) = 27, which is 8:42.

At this time, Xiao Zhang caught up with the cyclist and was 60 from the starting point (15+9), that is, Xiao Wang at 8:24.

Zhang's speed is 5 times that of the cyclist, so the cyclist's position when Zhang sets off is before Xiao Wang's 8 o'clock 24 5 = that is, Xiao Wang's position at 8 o'clock.

Xiao Wang and the cyclist are in the same position at 8:15, Xiao Wang's speed is 3 times that of the cyclist, and the cyclist should be in 3 * ( that is, 8 o'clock when the cyclist reaches the position of Xiao Wang at 8 o'clock.

It was 8:27:36 when Xiao Zhang set off from Adi.

In one city, some people were affected by the flu. In the first month, 20% of the population was infected with the flu, and only 80% were healthy.

In the next few months, 20% of patients recover and 20% of healthy people are infected with the disease.

So how many people are healthy at the end of the second month?

End of the first month: 20% are sick, 80% are healthy.

At the end of the second month: ** people: 20% x 20% = 4% people who are still healthy: (1-20%) x 80% = 64% So at the end of the second month, the healthy people are: 64% + 4% = 68%.

First of all, two digits plus two digits can only equal more than 100.

So, it can only be 1.

Second, the single-digit 1 pentagram results up to 1 in the 10th place

Therefore, the ten-digit 1 pentagram is at least equal to 9, and when the pentagram is 8, the single digit 1 8 has no carrying, and the equation does not hold.

So, Pentagram 9

Thus is introduced, 0