In the first year of high school, mathematics solving, speed finding, and the process of chasing sco

19.On the right, we can see a+3》0, and on the left, we can decompose aa-9=(a-3)(a+3), then the square of (a-3) * (2+3) in the root number is a-3"0, in summary, -3"a<3

20.I'll tell you next time I'm free, just ask the teacher if you can't, hehe.

19. The population after x years is a*(1+

20. Let the GDP in 2000 be a, then the GDP after x years will be b = a * (1 +, so that the GDP after x years will be b a = (1+ times) in 2000.

In fact, the two topics are of the same type, and both are the same average growth rate problem. The Enlu of both questions is the same, as follows: the average growth rate is a% in one year, the growth is (1+a%) after one year, then the growth is (1+a%)*1+a%) in two years, the growth is (1+a%)*1+a%)*1+a%), and the growth is (1+a%) n

So the result of question 19 is: y=a* (1+.)

The result of question 20 is: y=(1+ .).

The idea can refer to the above analysis, you can set the GDP after x years is y times of 2000, and 2000 GDP is a, then the GDP after x years is ay, ay = a * (1+, so y = (1+

(2)｜b｜=4

b+a and a-b are squared respectively and then subtracted to each other to get 4 a b =32 and finally b =4

The rest of the questions don't seem to be complete.

The total number - front row left and right adjacent - back row left and right adjacent + back row next to 3 seats.

First of all, there are different rows. Just 8*10*2=160

Then to the same row. First the front row.

Interpolation. Because it's 2 people. 10 digits. It is to insert two people into nine of the eight empty seats.

So. 9*8=72

Back. Distinguish whether it is on the same side or not.

Words on the other side. First it is 4*4*2=32

Same side. Also with the interpolation method. There are 4 digits left on one side. 2 people sit. That's 2 empty seats. There are 3 blanks to plug in. Two people are different. Sorting is required.

That is, a3 2=3*2=6

Because both sides are the same. So 6*2=12

Summing up. 160+72+32+12=276

1. Two people sit in separate rows: there should be 10 * 8 * 2 = 160 kinds 2. Two people sit in the front row: first assume that two people can be next to each other, there are 10 * 9 = 90 kinds; If the first person chooses a seat, the other person has two options 10*2=20, and then subtracts the two situations in which the first person is on the very edge, so there are 90-20+2=72 kinds.

3. Two people sit in the back row: a, two people sit on both sides of the bad seat, there are 4 * 4 * 2 = 32 kinds b, two people sit on the side of the bad seat, the same algorithm as the previous situation, there are 4 * 3-4 * 2 + 2 = 6, and because the situation on the left and right sides is the same, it has to be multiplied by 2, a total of 6 * 2 = 12 kinds.

To sum up, there are a total of 160 + 72 + 12 + 32 = 276 kinds.

1. Set the point n(x,0,0), then (x 4) 1 4=30, and the solution is x=9 or x= 1;

2. x y = 4 and x (y a) = 6 a, the common chord equation is ay = 1, (0, 0) the distance from this straight line is 1, then |1|/|a|=1, get a=1 (negative value rounded off).

Let n(x,0,0),|mn|^2=30.then (x-4) 2+(0-1) 2+(0-2) 2=30 then x=9, or x=-1

Solve the equation of the circle to remove the square term: the equation of the straight line where ay=1 chord length is located. Use the first center of the circle to the long distance of the chord (formula) d=1 |a|。d 2 + the square of the half chord length = the square of the radius, we get 1 a 2 + 3 = 4, so a = 1

（3）：2√3

18：-3<=a<=3

19: A times (1+ to the xth power.)

20: is 2000 (1+ x power multiples.)

18。(a-3)(a-9) 0, a+3 0 gets a -3 and both sides are positive, so 3-a 0, gets a 3, so -3 a 3.

19。Each year is a double of the previous year, a