A problem of mathematics integral in seventh grade volume 1 In a residential area, three x-wide paths were built on the rectangular lawn of length a and width b (with two paths and one path across). The lawn was divided into six equal parts. A, B and C were used to represent the area of lawn after road construction. (a is the length of the whole lawn, B is the width of the whole lawn)

A problem of mathematics integral in seventh grade volume 1 In a residential area, three x-wide paths were built on the rectangular lawn of length a and width b (with two paths and one path across). The lawn was divided into six equal parts. A, B and C were used to represent the area of lawn after road construction. (a is the length of the whole lawn, B is the width of the whole lawn)

b（a-1）
Because the width does not change, no matter how curved it is, it is like a straight one
The width is the same! So the two parts separated by the path can be put together!

Exercises 4.6, 1,2,3,4 on page 153 of PEP,

2. From the question, we can get 100 = 25 + 10t t = 7.5 3. After X years, the father's age is 4 times of his son's 40 + x = 4 (13 + x) x = - 4 4 4. The price is x yuan 80% x = 188 x = 235 6. The first number is x and the second number is (100-x) (x + 3) = (100-x) - 3 x = 47 100-x = 53

The world knows seven mathematics
The point m (a + 3,4-a) is known to be on the y-axis
(1) Find the value of a (2) translate the point m down three unit lengths, and then to the left two unit lengths, get the point n, and write out the coordinates of the point n
If the distance between point a (3a-7, - 1-A) and two coordinate axes is equal, and point a is in the fourth quadrant
(1) In terms of the value of a, the coordinates of point a
Given that point a (- 3,0) point B is on the coordinate axis and the distance between two points AB is 2, the coordinates of point B can be obtained
Given that point a (0,0) point B is on the coordinate axis and the distance between two points AB is 2, the coordinates of point B can be obtained

a=3;
n(-2,4);
a(-5,5),a=4;
B (- 1,0) or B (- 5,0);
B (2,0) or B (- 2,0) or B (0,2) or B (0, - 2)

At 24:3, - 6,4,10,

1: 3 × (4 + 10 - 6)
2: 3 × ((4 + 10) - 6)
3: 3 × (4 + (10 - 6))
4: 3 × (4 - 6 + 10)
5: 3 × ((4 - 6) + 10)
6: 3 × (4 - (6 - 10))
7: 3 × (10 + 4 - 6)
8: 3 × ((10 + 4) - 6)
9: 3 × (10 + (4 - 6))
10: (3 × (10 - 4)) + 6
Here are 10 ways. Hope to adopt ～～～

A and B start from ab at the same time. The speed ratio of a and B is 3:4. They meet at 9 kilometers away from the destination. How far is the distance between AB and B
2. How to change the end point of question 1 to the midpoint? I'll give you 300

I think it's easier to understand the equation
If the velocity of a is 3V, then the velocity of B is 4V, and the distance between the two places is s km
According to the theme
9÷3v = (s - 9)÷4v
S = 9 / 3 × 4 + 9 = 21 (km)
If the end point is changed to the middle point
Then (s / 2-9) △ 3V = (s / 2 + 9) △ 4V
S = 3 × 18 + 4 × 18 = 126 (km)

It is known that the length of the base of an isosceles triangle is 5cm, and the difference between the two parts of its circumference divided by the middle line of a waist is 2cm?

If the waist length is x, there are two cases
1. The solution of X + 1 / 2x + 2 = 5 + 1 / 2x is x = 3
2. X + 1 / 2x = 5 + 1 / 2x + 2 gives x = 7
So the waist length is 3cm or 7cm

Given X3 = y = Z2 ≠ 0, then XY + YZ + zxx2 − 3y2 + 4z2=______ ．

From X3 = y = Z2 ≠ 0, we can get: x = 3Y, z = 2Y, substituting: = XY + YZ + zxx2 − 3y2 + 4z2 = 3y2 + 2Y2 + 6y29y2 − 3y2 + 16y2 = 1122 = 12

Tree planting problem formula
Plant both ends, neither end, one end, one end

Tree planting problem 1 tree planting problem on non closed line can be divided into the following three cases: (1) if trees are to be planted at both ends of the non closed line, then: number of trees = number of segments + 1 = total length △ spacing + 1 total length = spacing × (number of trees-1) spacing = total length △ number of trees-1. (2) if trees are to be planted at one end of the non closed line, the other end

When the side length of a square increases by 3cm, its area increases by 39cm2. The side length of the square is ()
A. 5cmB. 6cmC. 8cmD. 10cm

Let the original side length of the square be x, then x2 + 39 = (x + 3) 2, the solution is x = 5, so a

Sequence {an} (n subscript), A1 = 8, A4 = 2, and satisfy an + 2 (n + 2 subscript) = 2An + 1 (n + 1 subscript) - an (n subscript). Let · Sn = ｜ a1 + ｜ A2 ｜
+An, Sn

This is the arithmetic sequence, A1 = 8, d = - 2, A5 = 0, A6 = - 2,. An = 10-2n
So when N5, Sn = (a1 + A2 + a3 + A4 + A5) - (A6 + A7 +. + an)
=20-(-2+10-2n)*(n-5)/2
=n^2 -9n +40