Solve a few math problems in the seventh grade, urgently

1.According to the title, 2a+b+2=7

3a+3b-9=3

The original system of equations can be reduced to.

2a+b=5①

a+b=4②

Get. 3a=1

a=1 3 substitute a=1 3.

1/3+b=4

b=13/3

So one-third 6=1 3*(1 3)+13 3*6+1 3*6=28 and 1 9=253 9

2.The speed of the person is xkm min, and the speed of the car is ykm min, according to the question.

6（y-x）=2（x+y）

6y-6x=2x+2y

4y=8xy=2x

The distance between each two vehicles is 2*(x+2x)=6x

t=6x/（2x）=3

A: There is a car every 3 minutes.

Question 1: Because x y = ax + by+ xy, substituting 2 1 = 7, where a and b are constants, then 7 = 2a + b + 2 * 1

In the same way, substituting (-3) 3=3 gives 3=(-3)a+3b+(-3)*3

Solution 2a+b+2*1=7

3）a+3b+（-3）*3=3

A system of equations, find the values of a and b a=1 3 b=13 3 substituting a=1 3 b=13 3 into x y=ax+by+xy to get x y=1 3x+13 3y+xy, and then substituting one-third 6 into x y=1 3x+13 3y+xy to get 1 3*1 3+13 3*6+1 3*6=253 9

The specific values will be calculated.

1. Solution: Bring x=2 y=1 into ax+by+xy=7 to get:

2a+b+2=7

2a+b=5 ①

Substituting x=-3 y=3 into ax+by+xy=7 to obtain.

3a+3b+(-9)=3

3a+3b=12, synthesis, to obtain the system of equations: 2a+b=53a+3b=12

3, get: 6a+3b=15, get: 9a=3

a=1 3 brings a equal to 1 3 into , gets.

b=13/3

So the original ax+by+xy=1 3x+13 3y+xy substitutes, x=1 3 y=6, and we get 1 3 1 3 +13 3 6+1 3 6=253 9

2: Solution:

1. a²+b²=(a+b)^2-2ab=232.1 nm = 10 9 m = 10 -7 cm, i.e.: 1 cm = 10 7 nm.

The diameter of 1 oxygen atom is (10 7) 10 8=nanometers.

3.Divide by seconds.

Hello, question 2 can't read clearly, is it a number to the power of 0? That's 1, and the power of zero of any number is 1, except for 0.

Question 1: A to the (2n) power.

x to the power of (2m+3).

There is a picture, let the small rectangle length is x, the height is 3x 5

x+3x/5+3x/5)^2-8*x(3x/5)=4(121*x^2)/25-(24*x^2)/5=4121x^2-120x^2=100

x^2=100

x=10mm

The small rectangle is 10mm long and 6mm high.

Set the length to x mm and the width to y mm

Figure 1 shows that 3x=5y

As can be seen from Figure 1, x+2=2y

The solution is x=10 y=6

It is 10mm long and 6mm wide

You can list binary equations.

Solution: Let the length of the small rectangle be x and the width be y.

If the parentheses can't be typed, it's an inline equation! ）

5x=3y2y-x=2

The solution is x=10

y=6A: The length of the small rectangle is 10mm, and the width is 6mm.

When the little dragon was puzzled, he found that 8 small rectangles of the same size could be put together into a large rectangle, as shown in Figure 1. Chen Ye saw it and said, I will try it too, so he put it together into a square as shown in Figure 2, leaving a gap in the middle for a square with a side length of 2mm Find the length and width of the small rectangle.

Solution, get: let the length be x mm, the width be y mm, then.

3x=5y，x+2y=2

The solution: x=10 11 y=6 11

So 10 11 mm long and 6 11 mm wide

Let the length be x mm and the width be y mm.

3x=5y，x+2y=2

The solution: x=10 11 y=6 11

So 10 11 mm long and 6 11 mm wide

Remove the brackets and extract the coefficients of the xy term, 3*(-3m)-1 2*7=0

9m-7/2=0

m=-7/18

The equation has to cancel out all the terms with xy, i.e., -9mxy, these two cancellations.

To offset, it is necessary to have a positive and a negative, that is, -9m= m=minus seven-eighteenths is not very clear, and the table is ......

Landlord, I'm the first, you must give it to me!!

The coefficients of the xy term are extracted.

9m+7/2)=0

m=-7/18

3（2x²-3mxy+y²）-x²+7xy-my²）

Multiply and open Merge the same terms 11 2x -(3m-7 2)xy + (1-m)y

Without xy, 3m-7 2 is equal to zero, 1-m is not equal to zero, m=7, 6

1 According to the substructure of the formula, and the value should be fixed, after the simplification of the knowledge, there is no a, then the first absolute value must become 4-5a, when the second absolute value is removed, it becomes 3a-1, the value of the number formula in this era is 3 (fixed), so 4-5a"0 and 1-3a, 0 get 1 3"a"4, 5

2 From the meaning of the question p=-2q into the equation when q 0 is known, the result is 4, and when q 0, the result is 6, so the result is 4 or 6

3 From the known conditions, 25 can only be split into 5 5 a b c d four numbers are either 2 positive and 2 negative or all positive or all negative and can only be 5 1 -1 -5 these four numbers and a" b "c" d 2 positive 2 negative can only ab be positive cd negative abcd is 5 5 -1 -1 or 5 1 -1 -5 all positive is 5 5 1 1 1 1 all negative is -1 -1 -5 -5 all 1 1 all negative when the result is 12

My answer I hope it helps you and I hope you are satisfied.

(3) Because 25=5*5, it is unlikely that a, b, c, and d are all positive numbers, because if they are all positive numbers, there are only two cases:

1. Both of the numbers are 5, but the requirements of the question are not equal to each other.

2. One of the numbers is 25, but in this way the other three numbers must all be 1, so negative integers should be taken into account. Because 5*5=25, consider turning a 5 into a negative, then it becomes 5*(-5)=-25, to make it positive, then multiply it by a -1, you get 5*(-5)*(1)=25, and finally as long as you multiply it by a 1, you can get four numbers multiplied by 5*(-5)*1*(-1)=25. Then the sum of these four numbers is, of course, 0.

2 I didn't understand it, I'm sorry.