Solve the problem, the first year of mathematics, there is an accurate answer, you will give a like,

1) Set the distance of 660 km after x hours.

408＋(72+96)x=660

408+168x=660

168x=252

x = two cars meet after x hours of express driving.

72+(72+96)x=408

72+168x=408

168x=336

x=23) Let the fast train meet the slow train after x hours.

72x+408) a 96x=60

Set x hours after 660m to get 72x 96x=660-408 Solution x= hours Let x hours meet, get (72 96)x=408-72 x=2 hours Set x hours, 96x-72x=60 hope, thank you.

1)660-408=252（km）

The distance between x is 252km

72x+96x=252

x = meet when x.

72（x+1）+96x=408

72x+72+96x=408

168x=336

x=23) is 60km apart when x

72x+96x=348

660-408 = 252 252 divided by 72 + 96 + time of encounter = 2x distance speed and.

Zhangjia 7x +11x-6

Wang 7x +11x-6-x+1= 7x +10x-5

Li 18x +27x-21-(7x +11x-6)-(7x +10x-5)=4x +6x-10

Withdrawal of shares of Jin Wangjia (7x +10x-5)x6=42x +60x-30

Li Jia (4x +6x-10)x10=40x +60x-100

42x +60x-30-(40x +60x-100)=2x +70>0

The two places are 12 kilometers apart, and A stays from A to B for 30 minutes and then returns from B to A. B travels from B to A, stays in A for 40 minutes, and then returns from A to B. It is known that the two people departed from A B at the same time, and 4 hours passed.

When they meet on their respective return journeys, if A's speed is faster than B's speed in kilometers per hour, find the speed of the two?

Let B's speed be a kilometer hour, then A's speed is a + kilometer hour 30 minutes = 1 2 hours, 40 minutes = 2 3 hours (4-2 3) a+(a+).

10/3a+7/2a+21/4=36

41/6a=123/4

a = kilometer-hour.

A's speed is kilometers per hour.

Solution: Let B's velocity be x kilometers per hour, then A is (kilometers per hour, according to the question:

Solution: x=x+Answer: A's velocity is 6 kilometers per hour, and B is kilometers per hour.

3+4+5+……17=150

Since the original formula is constant, after removing the absolute value, the constants of x add up to 0 and 150 2 = 75 = 17 + 16 + 15 + 14 + 13 so when 1 13< x<1 12 is a constant, and this constant is 17-(3-1)-5x2=5< p>

①.1 12 x 1 13 hours:

3x-1∣+∣4x-1∣+.12x-1∣+∣13-1∣+.17x-1∣

-(3x-1)-(4x-1)-.12x-1)]+13x-1)+(14x-1)+.17x-1)]

-75x+10)+(75x-5)=10-5=5;

When x=0: 3x-1 + 4x-1 +12x-1∣+∣13-1∣+.17x-1∣=15；

When 1 12 x 1 13 or x = 0, the absolute sum is a constant 5 or 15;

3+4+5+……17=150

150 2=75=I found that there is "Enthusiastic Netizens" below, and I will analyze the correct answer 5 or the minimum value of the sum of the absolute value.

Although it is an Olympiad, it is not necessarily extremely difficult.

Oh, that's a huge challenge.

First of all, we can be sure that this formula contains 15 absolute formulas (because 17-3+1=15).

The 0 point of each absolute formula is 1 3, 1 4, 1 5 ......1 17, zoom out in turn.

So when x increases, the absolute value of the rightmost equation will start to become positive first.

So we can know that, for example, when x=1 8, after these absolute values, we add the parts of the primary terms, and the result is.

3x-4x...7x+8x+9x...17x and if you want the whole equation to be equal to a constant, thenThe sum of the terms must be 0

When we understand what this problem is about, we can assume that when x=n, the sum of the coefficients of the terms = [-3-4-..n+(n+1)+.17]=(-3-n)(n-2)/2 + n+1+17)(17-n)/2=(-n²-n+6)/2 +(n²-n+306)/2=(-2n²-2n+312)/2

n²-n+156=0

The solution is n=12

So l12x-1l=-12x+1 and l13x-1l=13x-1x 1 12 and x 1 13

At this time there are 10 1's and 5 -1s in the constant part

Adding up equals 5 To sum up, the answer to this question is:

1 13 x 1 12 [Strict can be written in the form of intervals].

DBA Countless Solutions, 4, Right, Solid Point, In.

Because it is a one-dimensional equation, ax 2=0 and a=0.

x b-2=x，b=3。

x a+b=x 0+3=x 3。

Remember the thumbs up.

Because it's a one-time unitary, a=0 b=3