This has a regular math problem to solve

Figure 3 is used to analyze the pattern:

31 x 23 = 713

31: Two sets of horizontal lines, 3 and 1 respectively.

23: Two sets of horizontal lines on the left and right, 2 and 3 respectively.

Note: The high is on the top left and the low is on the bottom right.

The graph is then divided into 3 zones.

Zone 1: 6 intersections (representing hundreds).

Zone 2: 11 intersections (representing ten) Note that the intersection of the two diagonal parts is only sum.

Zone 3: 3 intersections (representing single digits).

So the sum is: 713

The same is true for the others.

Actually, this is a formula:

31 x 23 =（30+1）x （20+3） =（30x20）+[20x1）+（30x3）]+1x3）

Corresponding area: Zone 1 Zone 2 (two parts) Zone 3.

Number of points: 6 11 3

So the result is: 72384

The number represents the number of intersections, and 21x13 and so on represent the number of lines, such as 21 is two in the upper left corner, 1 in the lower right corner, 13 is one in the lower left corner, 3 is 3 in the upper right corner, and so on.

And if you want to draw the surface, you must first draw a line, 232, that is, from the upper left corner to the lower right corner, there are 2 strips, 3 strips, and 2 strips. 312 is 3 from the lower left corner to the upper right corner, 1 and 2. Then it's time to draw the curve with the intersection, like the one on the left.

Replace (n+1) n with a binomial as n n+1+......I can't remember exactly, but there are definitely the first two items, and then use the ratio method, i.e. n n (n + 1) n = n n (n n + 1 + ......Less than or equal to 1

So n n is less than or equal to (n+1) n

Equal when n=1.

Multiply the numbers in the bottom circle minus the numbers in the middle circle, which is equal to the number in the top circle, as follows. The first.

The second. And so on, the third.

So fill in 12 for the fourth.

15-10 5 fill in 5 in the middle

Actually, this question goes like this:

x - 1) (x - 1) , x n in the formula x cannot be equal to 1 , but the limit can be found.

When x ->1 the formula = 1 2

The answer is obvious.

To ask what is the answer to continue. Bring it in and you'll be out.

x = 0, formula = 1;

ps: I'm here to make soy sauce, ignore me)

PS: It occurred to me that x - 1) (x - 1) can be simplified:

x - 1)/(x - 1)

x - 1) (( x - 1)( x + 1)) = 1 ( x + 1), of course, x ≠ 1

But the equations are equal, and the limit values are equal over the common defined domain.

Just bring x = 1 into it and you will get 1 2, and you can save the derivative.

ABCD answers are all correct. The positive solution is: (root 1-1) 0, but the denominator is 0 and should not be. Therefore, there is no solution to this problem.

There is no answer, the root number above is subtracted by one, and the root number below is directly subtracted by one.

The number under the root number gradually decreases by one, and the denominator gradually decreases by one.

B Answer Because this is a number that gradually gets bigger.

It's just, there's no answer, isn't it equal to 0?

12All the twelve matches were put on a triangle.

Solution: If the length of the match is 1, how many possibilities are there to transform the problem into a triangle with a circumference of 12.

According to the law: the sum of the two sides is greater than the third side, and the difference between the two sides is less than the third side. There are three possibilities:

First of all, according to the figure, the sum of the number of quadrilaterals and triangles is the circumference of a large parallelogram or trapezoid composed of 1, 2, 3, 4, and 5, and then observe and find the law: when n is an odd number, its perimeter is 3n+5, and when n is an even number, its circumference is 3n+4, and then it can be solved according to the law

If the sum of the number of quadrilateral and triangular pieces of paper is 2, then the perimeter is: 2 5 = 10, if the sum of the number of quadrilateral and triangular pieces of paper is 3, then the perimeter is: 2 7 = 14, if the sum of the number of quadrilateral and triangular pieces of paper is 4, then the perimeter is: 2 8 = 16, if the sum of the number of quadrilateral and triangular pieces of paper is 5, then the perimeter is: 2 10 = 20, the law can be obtained: for each additional triangle, the perimeter increases by one side length, When the circumference increases by 2 sides for each additional quadrilateral, when n is an odd number, the circumference of n+12 quadrilaterals and n-12 triangles is: 2 (4+n-12 2+n-12 1)=3n+5, when n is even, by n2 quadrilaterals and n2 triangles, its perimeter is: 2 [4+(n2-1) 2+n2 1]=3n+4

The number of pieces of paper in a parallelogram and a triangle is n, is it all or all.

Hehe, the printing of this question is not very standardized.

Actually, it's like this:

Singular terms: 4, 7, () 19

Even: 6, 12, 18, 24

It's very regular!

In parentheses is 12, choose b.

The odd terms are preceded and followed by 3, 5, and 7 increments, in that order

d, which is 12 larger than the previous counterpart

Assuming the 7m student is like this, then it is followed by 6n 1 7m 6n 1 101

7m＋6n＝101

Obviously m can only be odd.

(m,n) (5,11) or (11,4) has only two sets of solutions, so only two students are like this, students 35 and 77.

Because it's an even number, there will definitely be two people counting each other.

The 7 that can be reported within 100 is 7 14 21 28 35 "42" 49 56 63 70 77 "84" 91 98

Within 100, the 6 that can be reported is 6 12 18 24 30 36 "42" 48 54 60 66 72 78 "84" 90 96

That's it.

Report 7 is a multiple of the 7th person from left to right Report 6 is a multiple of the 6th person from right to left If it is the same person, then it should be satisfied 6 * m + 7 * n = 101 m and n are natural numbers m is taken from 0 to 16 and the corresponding n is true if it is satisfied is a natural number so when m is satisfied and is the 24th and 84th people from left to right.

The equal sign of each equation is followed by the last digit of the previous equation plus ......form.

And so on.

I don't know how to ask me.