Ask for a detailed analysis of two junior high school math problems, urgent!!

A set of monthly growth rates are stated.

500（1 + x）2 = 720

The solution is x = 12

One (1-x 2)-2 2bx 1 + 2 = 0 simplified, we have:

a + 1) 2 - 2 2b) x + (a + 1) = 0, so x1 * 2 = (1 +) (1).

x1 + x2 = (2 2b) (1-a).

So x1 2 + 2 times 2 = (x1 + x2)2 - is 2x1x2 (8b 2-1-2-2a) (1-a) 2 = 12 oriented simplification, we also: 13a 2 - 22a +13 = 8b 2 no more. So, if it weren't for some misunderstanding on my part of the expression of the owners, it would be a problem to say wrong.

According to the Pythagorean theorem, then CB represents.

One: B: C can be ordered.

1.Add x dollars. Ticket revenue per day is y.

40+x）*（380-2x）=y

Q1: When y=2400, x=40. **Should be set at 40 + 40 = yuan. Question 2: As can be seen from the question, when y is maximum, x is the amount that should be increased.

y=The meaning of the question is that when y is maximum, x is the amount that should be increased.

y=15200+300x-x ·· When x=150, y is maximum. So the ticket should be increased by 150 yuan.

2.Connect 3 points to form a triangle and make a vertical bisector of three sides, and the three lines will intersect at one point, and that point is it.

The first question, the first question, set the ticket ** should be set as x yuan, then according to the meaning of the question there is x*(380-2(x-40)) = 24000

Let's figure it out for ourselves.

The second question is the first question to be calculated, and then the X-40 is the increased price.

Question 2. Connect the three points to form a triangle, and then make three and a half vertical bisectors each, and the intersection of the three is the herdsmen's settlement.

It's an inequality problem.

We set an increase of X yuan for tickets, which can make the daily ticket revenue reach 24,000 yuan.

And then we have:

The ticket price is: 40+x

yuan, the number of visitors is.

2x yuan. The total revenue is: Fare.

Times. Number of visitors.

i.e. (40+x)*(380.)

2x）》=24000

150x solution: x

That is, the fare is set at.

Between. Fare. Increase by 44

Between. As for the second question, hypothetical. ab

C points are three. 3 grazing spots.

Connect. AB, BC, respectively as line segments. ab.

Median line. m，bc

median line. n, straight line. m with.

The intersection of n. It is. Pastoralist settlements.

The principle is the nature of the median line.

I told you that it was 24000=[40+x] [380-2x] and you realized it again. Look, everybody's talking about it now.

What you set here is to increase x yuan, and it is calculated to be 40 and 110, and you only need to add back the original money, which is the final answer.

You have to draw a picture of the second question, I know if you have 3 points connected into a circle or a triangle.

(1) Admission fees shall be set as:

X yuan, the number of people is.

380-(x-40)*2, according to the title.

x*(380-(x-40)*2)=24000. xx2=150 or rather.

x《=150

2) Suppose the ticket should be increased by x yuan, and the number of people is.

380-2x, depending on the title.

x+40)(380-2x)=24000

Solution. x1=x2

1.Set the ticket price to 40 x RMB, then there is:

40＋x)*(380-2x)=24000x^2-150x+4400=0

x = 40 or 110 (minus irrelevance).

Tickets** should be set at $80 and tickets should be increased by $40.

2.According to the nature of the outer center of the triangle, the point sought is the outer center of the triangle that connects the three grazing points, and the outer center is the herdsman's settlement. Here's how:

Connect the three grazing points to form a triangle (of course, it is impossible to achieve the requirements of the problem if the three grazing points are in the same straight line).

Make the perpendicular bisector of both sides, and their intersection is what you want!

Question 1. Equation.

24000=【40+x】【380-2x】

This setting increases by x yuan.

2x fewer people per day.

You add that set to increase by x.

Reduce 3 molecules per day by 5 people.

1 Extend ab, cd and E, E=30, ae=6, be=2*root number 3, ab=6-2*root number 3

2 In the right triangle CDF, (to be clear that CF=BC), CD:CF=4:5, then DF:CF=3:5, so, af:AD=2:5, you might as well set BE:AB=X:4, in the right triangle AEF, (* is the multiplier sign) X*X=(4-X)*(4-X)+2*2, the solution is X=, in the right triangle ECF, EF=BE, CF=BC, COS ECF=5 root number (25 + root number 5

Extend the intersection of DC and AB E

d=90°，∠a=60°，ad=3

ae=ad/cosa=6 ∠e=30°

cba=90°

cbe=180°-∠cba=90°

and e=30° bc=2

be=bccote=2, root, number 3

ab = ae-be = 6-2 root number 3

Let ab=4a, bc=5a

then cf = bc = 5a cd = ab = 4a

d=90°df=3aaf=ad-df=2a

efc=∠b=90°

afe+∠cfb=90°

a=90°aef+∠afe=90°

cfb=∠aef

In AEF and BFC.

a=∠b=90°

aef=∠bfc

aef∽△bfc

ef fc = af cd = 2a 4a = 1 2 cos ecf = 2 root number 5 5

1 Extend AB and CD to H

From the meaning of the question, h=30, use the trigonometric function of 30 angles to find ah bh, and then calculate ab

Let me think about the second one.

So when x=3, y is the largest, and the value is 6400

yi=[5000-(x-100)*10]*x=-10x^2+6000*x(100≤x≤250)

y1=3500*x(250＜x)

y2=5000*

So you can buy up to 400 pieces.

I'm raising people to the fourth level, how can I rise up quickly, I'll solve your math problems.

The first question is wrong, it should be that CE is perpendicular to AB, because CE is perpendicular to AB, and BE=CE, so the triangle BCE is an isosceles right triangle, so B=45 degrees, and because B- A is equal to 30°, A=15 degrees.

Question 2: Suppose Group A produces x units a day, then there are:

6 x 5=(4*x+300+100) 4 solution x=500, group B produces 6x 5=600 per day, so group A produces 500 per day, and group B produces 600 per dayQuestion 3: Suppose x 2-pointers are shot, then there is equation:

2*3+2*x+(14-2-x)*1=22 to get x=4

So he made four 2-pointers.

1：∠b=120° ∠a=90°

2: Let A produce x a day, and B produce y a day.

From the known x:y=5:6, there is {5y=6x,300+4x+100=4y, and x=625,y=750

3: If you hit x two-pointers, you hit 12-x free throws.

2*3+2x+12-x=22 gives x=4, then 12-x=8. So he made four two-pointers and eight free throws.

Set A to produce x B y.

then x+5x=5y

300+4x=4y-100

300+4x=

x=250y=300

Yes, if a≠0 is right.

Multiply the equation and then translate, x can be eliminated, so it is not a one-dimensional quadratic equation The equation results in 0, and the numerator of the fraction is not 0, so it is satisfied only if the denominator of the fraction is 0. then 3x -x is meaningless.

If only "px -3x + p-q=0 is a quadratic equation about x", then it means that x can be solved or not, then p≠0 is sufficient.

The first question should be A, analysis: The concept of a quadratic equation is: an integer equation that contains only one unknown number and the highest order of the unknown number is 2.

So 3x +7=0 is; ax +bx+c=0 coefficient a is not equal to 0, but he does not add this restriction, so it is not; (x-2)(x+5)=x-1 remove the parentheses and move the back x term to offset without x terms, so it's not; 3x -5/x = 0 denominator contains unknowns, not an integer equation, so it is not.

The second question should be C, for the same reason as the first question

9 teams. If there are x teams, then there is 3x+2(x-6)=33, and the solution is x=9

Set to pass x people.

3x+3x·x+3x=864×50%

x=12 passed 3x people in the first round 3x people in the second round and wore x people on average in the second round The neighborhood committee also passed in the second round So the second round passed a total of 3x·x+3x people.

Let the taproot of this plant grow a branch.

1+a+axa/3+2axa/6=109a=12

Analysis: This problem can be set up that an average of one computer in each round of infection will infect x computers, then a total of (1+x) will be infected after the first round, and a total of (1+x)+x(1+x) that is, (1+x) 2 will be infected after the second round, and the value of x can be found by using the equation, and a total of (1+x)3 will be infected after 3 rounds, and the number can be judged by comparing the size of the number with 700

Answer: Solution: If each computer in each round of infection will infect x computers on average, according to the problem: 1+x+(1+x)x=81, and (1+x)2=81, then x+1=9 or x+1=-9, x1=8, x2=-10 (rounded), (1+x)3=(1+8)3=729 700

A: In each round of infection, an average of 8 computers will be infected per computer, and after 3 rounds of infection, more than 700 computers will be infected.

1) If the number of infections per person a day is x, Xiao Ming alone at the beginning; The number of newly infected people on the first day was x, and the number of co-infected people as of the first day was x+1;The number of newly infected people on the second day is (x+1)*x, and the number of co-infected people is (x+1)*x+x+1=225 on the second day, so x=14, then the first communicator has a total of 28 people in two days. (2) The number of newly infected people on the third day is 225*(14-5)=2025, the number of co-infected people as of the third day is 2025+225=2250, the number of newly infected people on the fourth day is 2250*(14-5-5)=9000, and the number of co-infected people as of the fourth day is 9000+2250=11250. The above is a detailed analysis.

If you are familiar with the number series, you can simplify the formula.