A math problem which I think is quite difficult .

Set the original price of x yuan, then the current price x-3, the original audience m, now.

So (1+1 5)x*m=(x-3)*

x=15 yuan.

Set the original price to x and the number of viewers to y, according to the title.

1+1 5)xy=(x-3)*(1+1 2)y. x=15

The original price is 15 yuan.

The original price of the movie ticket is X yuan, and the number of spectators is Y.

x-3）*（3/2）y*5/6=xy

The solution is x=15

The original price is 15 yuan.

Question 1: Except for one, the square of all prime numbers is e.g. 2 2=4;3^2=9，5^2=25;7^2=49...Wait a minute.

Question 2: If the number of apples in B is x, then A is 2x, so x+2x=3x>(55-10)=45; Find the maximum.

Solution: x=14, so A gets 28, B gets 14, and C gets 13

The first number is nine, and the factors are.

One, three, nine! The second A twenty-eighth, B fourteen, C thirteen!

The factors are 1, 2, 4

2, A 28

B 14C 13

1) The 3 factors in the question should not include 1 and itself, otherwise the square number of the prime number will meet the condition!

If it does not include 1 and itself, 16 is satisfied.

2) The sum of A and B is 55 minus more than 10, and the result is less than 45 and can be divisible by three;

42 satisfies the conditions, calculated A is 28, B is 14, C is 13, and the question is satisfied.

Question 1:

This integer is a perfectly squared number, and its arithmetic square root must be prime, e.g. 4, 9, 25, etc.

This problem can be expanded to read: If a positive integer has odd factors, then the positive integer must be a perfect square number.

Question 2: A 28.

B 14.

C 13.

The first question is to ask what is the smallest integer? This number is 4, otherwise there are infinite, such as 9, 25, 49, 121....Question 2: Let B be x, A be 2x, and C be y.

Then there is 2x+x+y=55, y=55-3x, because y>10, so 55-3x>10, x<15. When x=14, y=13, when x=13, y=9 is less than 10, so x can only take 14, at this time, A28, B, 14C, 13

Let's have a standard answer.

1) Let the speed of the elder brother be x (m seconds), the speed of the younger brother be y (m s), and the total length of the playground is s (m), then the conditions can be obtained s (x+y)=25 (s), 20s (x-y)=25*60 (s).

Divide the two equations and subtract s to get x y=2

2) From x=2y, we get s (2y+y)=25 (seconds), so s y=25*3=75 (seconds).

That is, it takes 75 (seconds) for Xiao Ming to run a lap, so Xiao Ming ran 25 * 60 75 = 20 laps In addition, it needs to be mentioned that there is a logical error in this question, because my brother wants to run one more lap to catch up with Xiao Ming, why run 20 laps? Detailed enough.

25*60 25=60 25 seconds together for one lap, so 25 minutes for a total of 60 laps.

60-20) 2=20 laps, Xiao Ming.

60-20 = 40 laps, brother.

40 20 = 2 times.

Let ab=c, ac=b, cb=a midline. a

The length of the midline of the side is ma=(1 2) 2b +2c -a )

Bring in to get bc=9

Let bc=2x, find the angle cad according to ac,ad,x, and find the angle bad according to ab,ad,x, and the angle bad+angle cad=angle bac.

Find the angle bac according to ab, ac, 2x.

If the two angles bac are equal, you can find x, and you can find bc.

Ok, do the math yourself.

Too troublesome bird ......Use the cosine theorem twice.

The result was 9

Extending AD, the parallel line of AB CE crosses the extension line of AD at E, then D is the midpoint of AE CE 2=AC2+AE 2-2AC*AACos CAE can find COS CAE

CD 2 = AD 2 + AC 2 - 2 AC Cae to CD = 9 2 BC = 9

Make use of the triangle midline theorem.

That is, the sum of the squares of the two waists is doubled, which is equal to the sum of the square of the bottom edge and the square of the middle line of the side 4 times) to obtain (16 + 49) * 2 = 4 * 7 2 * 7 2 + bc square.

The solution yields bc=9