Math Problem Solving

a^2=ab^3=c

b*d=d10b+a)*e=100a+10d+e(10a+b)*(10c+e)=?

a=0 or 1, d=0 or b=1, and a, b, c, d, e are different and all are 0-9if a=0,b=1,then c=1,rounded.

In the same way, only a=1 and d=0

b 3 = c, then b = 2, c = 8

10BE+E=100+E, that is, BE=10, so E=5 original formula = 100AC+10AE+10BC+BE=100C+10E+10BC+BE

a*1000+d*100+b*10+d

Answer: 8 8=8 Inference: 8=1

9 3=3 inferences: 3=0

9 9 9=5 inferences: 9=2 5=8

93+8) 7=837 Inferred: 7=5

Answer: 89 57 = 12 85 = 1020

8 8=8 inferences: 8=1

9 3=3 inferences: 3=0

9 9 9=5 inferences: 9=2 5=8

93+8) 7=837 Inferred: 7=5

Solution: 89 57 = 12x85 = 1020

Summary. a1=2,a2=4a(n+2)-an=(-1) n+3, i.e. a(n+2)=an+(-1) n+3a3=a1-1+3=4a4=a2+1+3=8a5=a3-1+3=6a6=a4+1+3=12The odd term is an equal series of differences with a tolerance of 2, the even term is an equal series of equal numbers with a tolerance of 4, the sum of the first 20 odd terms is 10 (2+2+(10-1) 2) 2=110, the sum of the first 20 odd terms is 10 (4+4+(10-1). 4) 2 = 220 for the first 20 terms and the sum is 220 + 110 = 330

Question 10. Why is my process wrong.

a1 = 2, collapse of the celery a2 = 4a(n+2)-an=(-1) n+3, i.e. a(n+2)=an+(-1) n+3a3=a1-1+3=4a4=a2+1+3=8a5=a3-1+3=6a6=a4+1+3=12The odd term is an equal difference series with a tolerance of 2, the even number of terms is a series of equal differences with a tolerance of 4, and the sum of the first 20 odd terms is 10 (2+2+(10-1) 2) 2=110The sum of the first 20 odd terms is 10 (4+4+(10-1) 4) 2=220, the sum of the first 20 items is 220+110=330

Dear, I calculated that the result was also 330

Dear, you are just like me, you should be right.

Dear, I calculated it again, and the result should be no problem.

Thank. Not being right with 330 is wrong.

Dear, I listed the top 20 items, and finally found the top 20 items and 330, there should be no problem, you can take a look.

Summary. Equation problems generally only need to find the equation relationship that exists in the problem, rely on the equation column equation, and then solve the equation.

Hello, can you send me the title?

I have to draw a picture.

Solution: It turns out that warehouse B has 11x tons, so A has 12x tons in warehouse for example. So (11x+24)-12x=1 9 (11x+24)11x+24-12x=11 9x+8 3-x+24=11 9x+8 3-20 9x=-64 3x=64 3 9 20x=(18 3) 5x=ton bucket racket.

Zhiqi. Can you draw a line diagram?

Seems to be the same. Yes, I think that's the same thing.

So I'll check it again.

I checked and mine was correct.

Answer: Warehouse B originally stored tons of fertilizer.

Equation problems generally only need to find the equation relationship that exists in the problem sail and slow, rely on the equation state modulo equation, and then solve the equation can be Lao Fu.

That's weird.,Why is it in the homework gang??

The homework gang miscalculated.

In the end, the result was wrong, I calculated according to the homework gang just now, and the last step on the homework gang was wrong.

At the time of departure, the speed ratio of A and B was 5:4, and when they met for the first time, A and B traveled the whole journey of 5 9 and 4 9 respectively.

After the first encounter, A's speed increased by 20 and B's speed increased by 20, at this time, the speed ratio of A and B became 5 (1+20%) 4 (1+20%) = 5:4;

When A arrives at place B, A travels 4 9 again, and in the same time, B travels (5 9) (5 4) = 4 9, so the distance between A and B is 10 (1-4 9-4 9) = 90 km.

At the time of departure, the speed ratio of A and B was 5 4, (equivalent to 20% faster than B), and when they met for the first time, A and B traveled the whole 5 9 and 4 9 respectively.

After the first encounter, A's speed decreases by 20 and B's speed increases by 20, (which is equivalent to the same speed as before the encounter).

This means that when A arrives at place B (after walking the remaining 4 9), B has also traveled 4 9, and B has traveled 8 9 of the total distance before and after the meeting

So, the distance between A and B is 10 (1 9)=90 (km).

There are two pieces of paper, one decimeter and the other decimeter long, pasted into a long strip of paper with a total length of 3 decimeters. How many decimeters is used for pasting?

At the time of departure, the speed ratio of A and B is 5:4, which is set to 5x, and when 4x meets for the first time, A and B travel the whole journey of 5 9 and 4 9 respectively.

After the first encounter, A's speed increases by 20, and B's speed increases by 20, at this time, the speed of two people A and B becomes 5x(1-20%) =4x,4x(1+20%)=;

When A arrives at place B, A travels 4 9 again, and in the same time, B travels ((4 9) 4x)* Therefore, the distance between A and B is 10 ((5 9)-(8 15)) = 450 km.

20 times 5 plus 20 times 4.

The original velocity ratio is 3:2=27:18

Now the speed ratio is::13

A walked 27 + 18 = 45 (exactly a whole journey), and B walked 31 parts, which is still 14 proportion points (that is, 14 kilometers).

So 1 scale point is 1 kilometer.

ab The distance between the two places x

[5*(1-20%)]/[4*(1+20%)]4/9)*x]/[(5/9)*x-10]=4/6*4x/9=5*(5x/9-10)

24x/9=25x/9-50

24x=25x-50*9

x=50*9=450

The distance between the two places is 450 km.

In addition, the 20 yuan hidden by the waiter is 290 yuan

270 + 20 = 290 (yuan).

20 here is wrong, 270 refers to the total amount of money paid by the three guests, and 20 is included in this 270 yuan, which is obviously not sorted out the logical relationship.

"Three passengers, one person 90, three people 270 plus waiter 20" is wrong, so the 10 yuan is missing.

According to the question, after the refund, each person is paid 80 yuan, plus 10 yuan per person refunded, a total of 90 yuan.

In fact, after the boss discounted 50 yuan, they should pay 250 3= each

You can see what the problem is wrong.

This kind of problem is a logical thinking problem.

The 20 yuan taken by the waiter was also from the three people who stayed in the hotel. So the total amount of money paid by these three people is: 250 yuan from the boss plus 20 yuan taken by the waiter, a total of 270 yuan.

And the 300 yuan was returned like this: the boss's 250 yuan plus the 20 yuan taken by the waiter, plus the 30 yuan that the waiter retrieved from the three of them, so that the 300 yuan was calculated.