Ask a simple math problem for beginner 1 to process

1-x=0, x=1, 2x-1=0, x=1 2

This question is divided into x 1, 1 x 1 2, x 1 2.

When x 1, 1-x 0, |1-x|=x-1. 2x-1＞0,|2x-1|=2x-1.

1-x|+|2x-1|=3 becomes x-1+2x-1=3x=5 3 (in line with the title, i.e. x=5 3 is in the range of x 1), when 1 x 1 2, 1-x 0, |1-x|=1-x. 2x-1≥0,|2x-1|=2x-1.

1-x|+|2x-1|=3 becomes 1-x+2x-1=3x=3 (does not fit the topic, i.e. x=3 is not in the range of 1 x 1 2, rounded off), when x 1 2, 1-x 0, |1-x|=1-x. 2x-1＜0,|2x-1|=1-2x.

1-x|+|2x-1|=3 becomes 1-x+1-2x=3x=-1 3 (in line with the title, i.e. x=-1 3 is in the range of x 1 2) so x=5 3 or x=-1 3.

Above, you are wrong, the value range of x, the first number is the maximum is 3, the minimum is -3, the maximum x is 4, the minimum is -2, the second number, the maximum is 3, the minimum is -3, the maximum x is 2, the minimum is -1, in the two sets of value ranges, find their common number, that is, both are suitable, find out that it is -1, x=-1 do not believe in the calculation. Above, you said it is 3, I will check it for you, 1-3=-2, the absolute value of -2 is 2, the second is 5, the absolute value of 5 is 5, 2+5 is equal to 7, you are wrong.

Because the sum of the inner angles of the polygon is (n-2)*180

The equation is: (n-2)*180=360n=4

4 such as squares

Degrees of each inner angle of a regular n-side: 180(n-2) n

The number of degrees of each outer angle of the regular n-side: 360 n The sum of the outer angles of the regular n-sided is 360 degrees).

X tons of water were used.

10*11+2x+

x=18 It's the first time I've typed such a complete question... Be sure to give it to me!

14, solution: let the pumping water of each pumping machine be x minutes, the river water continues to gush out, assuming that the amount of water gushing out per minute is y, and set the pumping machine before pumping, the volume of water in the depression is v, then it is known by the problem, and the following equation is obtained:

v=40*（2x-y）

v=16*（4x-y）

Solve the system of equations, get.

x=3y/2

v = 80y set up pumping machine k, can pump the water within 10 minutes, then.

80y=10*（kx-y）=10*（k*3y/2-y）=5*（3k-2）y

16=3k-2

k=6A: If you want to pump water in 10 minutes, then you need at least 6 pumps.

15. Solution: Let the teacher be t, and the student s1, s2

The teacher first takes S1 to ride a motorcycle, and S2 walks there, setting the time to T1

Before arriving at the museum, the teacher went back to pick up S2 and asked S1 to walk to the museum, and the time from the beginning of the return trip to the time when the teacher met S1 was T2

Finally, the teacher took S2 to the museum on a motorcycle, and S1 also arrived at the museum at T3

The equation is 20t1+5(t2+t3)=33

30t2+5t1+5(t2+t3)=33

5(t1+t2)+20t3=33

The solution is t1 = t3 = hours.

t2 = hours.

t1 + t2 + t3 = 3 hours.

It did not take more than 3 hours to get the result.

16, solution: Ming Ming * Bai Bai = Ming * 11 * White * 11 = Ming * White * 121 = Clear Chu = Clear 0 Chu * 11

So clear 0 is divisible by 11.

Divisible by 11 is the sum of odd digits minus the sum of even digits which is a multiple of 11.

And Qing and Chu are both single digits, so Qing + Chu can only be equal to 11

So Qing 0 Chu could be 209 308 407 506 605 704 803 902

Divide these numbers by 11 to get 19, 28, 37, 46, 55, 64, 73, 82, respectively

So the clear* white is 19 28 37 46 55 64 73 82

Both Ming and White are single digits, so exclude 19 37 46 55 73 82

So the clear* white is equal to 28 64

So Ming and White are 4 and 7 or 8 and 8 or 7 and 4 respectively

So the equation is 44*77=3388 or 77*44=3388 or 88*88=7744

So it's clear that it's 4738 or 7438 or 8874

14 Pumping rate x escape rate y Total water volume z

Then (2x-y)40=(4x-y)16=z gives 2x=3y, z=80y

z=(ax-y)10, and the solution is a=6

15. In the scheme design, the teacher first sends the student to a place that he can reach on time on foot, and then goes back to pick up the second student.

Process 1: The teacher first sends the student to a place where he can reach on time, with a time of a20a+5(3-a)=33, and a = 6 5 Process 2: The teacher picks up the student 2, and the time is b

Since 20a = 24

25b+5(a+b)=24, solution b=3 5 process three: the teacher sends the student three, and the time is c

Since 20a-25b = 9

Then 20c = 33-9 and c = 6 5

a+b+c=3, the scheme is established.

16 Ming Ming * Bai Bai = Qing Qing, indicating that "Qing Qing" is a multiple of 11 * 11, that is to say, "Qing 0 Chu" is a multiple of 11, then there must be "Qing" + "Chu" = 11, considering that "Qing 0 Chu" = "Ming" * "White" * 11, "Ming, Bai, Qing, Chu" means 4 different numbers, you can try out "clear and clear" = 4738 or 7438 one by one

Question 14: Suppose the amount of water gushing out per minute is x, the amount of water pumped by the pump per minute is y, and the original amount of water is z, then the equation can be derived according to the question.

40*x+z=2*40*y...1）16*x+z=4*16*y...2) Calculated: y=

Bringing Eq. (3) into (1) and (2) gives Z=80x....4) Therefore, it is assumed that the water is pumped within 10 minutes, and M pumps are required according to the title: 10*x+z=m*10*y

Bring y= into the above equation.

m=6So, to pump the water in 10 minutes, you need at least 6 pumps.

If there are x peanuts in total, the number of peanuts left after the first day is.

x-x 10-1 10=9x 10-1 10 The number of peanuts left after the next day is.

9x/10-1/10-(9x/10-1/10)/9-1/9=9x/10-1/10-x/10+1/90-1/9=8x/10-2/10

The number of peanuts left after the third day is:

8x/10-2/10-(8x/10-2/10)/8-1/8=8x/10-2/10-x/10+1/40-1/8=7x/10-3/10

The number of peanuts left after the ninth day is:

x 10-9 10=10 The solution gives x=109

So there were 109 peanuts in total.

Let the total number of peanuts be x, which is obtained from the equation of the question column:

1-（1/10)²]x*[1-(1/9) ²1-(1/8) ²1-(1/7) ²1-(1/6) ²1-(1/5) ²1-(1/4) ²1-(1/3) ²1-(1/2) ²10

Solve the equation to get the right answer!

1.Solution: Set up x pieces to prepare.

Left, center, right.

x÷3 x÷3 x÷3

x÷3-10 x÷3+10 x÷3

x÷3-10 x÷3+15 x÷3-5

I don't know what the format required by your school is), as you can see from the above**, there are (x 3+15) chess pieces in the middle at this time. As for the knowledge of elementary integers, you can find out how to explain them in the book.

2.Solution: Set up x pieces to prepare.

Left, center, right.

x÷3 x÷3 x÷3

x÷3-a x÷3+a x÷3

x÷3-a x÷3+a+b x÷3-b

Answer: The number of pieces left in the middle pile is (x 3+a+b).

Set: The pieces prepared by each person are 3x (x 12).

Divide into three aliquots ab c x

a＝x－10 c＝x＋10

b＝x－5 c＝x＋10＋5

a x 10 x 10 c x 10 5 (x 10) 25 check: a b c 2x 20 x 5 25 3x Answer: There are 25 pieces in the pile

Replace 10 in instruction with a, and 5 in instruction with b a x a c x a

b＝x－b c＝x＋a＋b

a x a x a c x a b (x a) 2a b check: a b c 2x 2a x b 2a b 3x Answer: There are 2a b pieces in the middle pile.

1.Left, center, right.

First time x x x

The second x-10 x+10 x

The third x-10 x+15 x-5

Fourth, 2(x-10) x+15-(x-10)=25 x-5x+15-(x-10)=25

At this point, there are 25 pieces in the middle pile.

2.Left, center, right.

First time x x x

Second x-a x+a x

3rd x-a x+a+b x-b

Fourth 2(x-a) x+a+b-(x-a) x-bx+a+b-(x-a)=2a+b

At this time, there are 2A+B pieces in the middle pile.

After setting the number of three equal parts to x, the change law of the three piles is easy to obtain according to the operation steps as follows:

x x xx-a x+a x

x-a x+a+b x-b

2(x-a) 2a+b x-b

So, when a = 10 b = 5, there are 25 grains left in the middle pile.

The number of pieces in each pile after each move.

n n nn-10 n+10 n

After the middle heap takes out n-10, the remainder: n+10-(n-10) = 20 After the left heap is put into n-10: n-10+(n-10)=2n-20 final: 2n-20 20 n

10 is changed to the letter a.

n n nn-a n+a n

After the middle heap is taken out of n-a, the remaining: n+a-(n-a) = 2a, after the left heap is put into n-a: n-a+(n-a)=2n-2a final: 2n-2a 2a n, there is no 5

Let each pile be x at the beginning, then after 2 small steps, each pile is x-10, x+10, x, then after the third step, each pile has 2 (x-10), x+10-(x-10), x, and x, then the number of pieces in the middle pile is obvious

Your title is not accurate.

Isn't there 20 left in the heap

1. Let there be a total of 3n pieces, and n is greater than 12

then at first each was.

n n nn-10 n+10 nn-10 n+10-(n-10) n+(n-10)

Apparently there are 20 in the pile.

2. **5?

Unclear ......

There are 20 in the middle pile.

Let's say that there are x pieces in each of the 3 piles.

After the second step, it becomes x-10 x+10 x, and after the third step, it becomes 2(x-10) x+10-(x-10) x, so the middle pile can be directly calculated as having 20 pieces.

I didn't see the number 5 in** so I can't do it... Hehe.

The time to go up is s v1 and the time to go down is s v2, so the total time is s v1 + s v2

Average speed = total distance Total time = 2s (s v1+s v2) = 2 (1 v1 + 1 v2) = 2v1v2 (v1+v2) Wrong......Student A moved C bricks in hour B, that is, every hour C B block C moved A hour at the same speed, it is A*C B block, but it must be an integer.

Average speed = total distance Total time = 2s [s v1 + s v2].

2v1v2/[v1+v2]

Ascent time s v1 descent time s v2

The draw speed is s (s v1 + s v2).

Simplification yields the average speed (v1*v2) (v1+v2).