A few math problems admit that I didn t learn this lesson well, so please help me urgently!

1 xy(x^2y^2+2xy+1)=xy(xy+1)^22 4x^2-1= (2x+1)(2x-1)3 b4 a

1.Decomposition Factor:

m^2-n^2+2m-2n =m^2+2m+1-n^2-2n-1=(m+1)^2-(n+1)^2=(m+n+2)(m-n)

2.When a = root number 2 and b = 1, find the value of a 4-a 2b 2 of a 2-ab.

The numerator is 2-root number 2, and the denominator is 4-2=2, so it is one minus two-part root number two.

It seems that you should be a primary school student in the first year of junior high school, right, your age is the easiest to change your life's character, and it will determine your future, so work hard, you won't do these questions now, but when you take the college entrance examination or as a college student, if you can't see the final result in your head for three seconds, then you may be a loser, learning, should be the main theme of your stage, remember.

2.(2x+1)(2x-1)

1-1=m^2+2m+1-(n^2+2n+1)=(m+1)^2-(n+1)^2

m-1-n+1)(m+1+n+1)=(m-n)(m+n+2)2.Original formula = a(a-b) a 2(a 2-b 2) = 1 a(a + b) = 1 a 2 + ab = 1 2 + root number 2

2 - root number 2) 2 = 1 - (root number 2) 2

Option 1 is equivalent to a choice (mail order of less than 10 books (including 10 books) will add a mail order fee of 3 yuan) Option 2 is equivalent to a choice (mail order of more than 10 books (excluding 10 books) will add a mail order fee of 15% of the book price).

1) Scheme 1: (, 390+4=394 Scheme 2 You do the math.

3024 (167-83) = 36 days.

It is equivalent to doing 167 meters a day in the past, and now repairing 167-83 meters a day, and then dividing the total length by what you do every day.

The original plan was 3024 167 = days to complete.

And in 3024 94 = days.

It will take X days to complete.

x*（167-83）=3024

So x=36

The distance from the township school to the county seat is s

The speed of the car is 6v

The speed of walking is v

The normal time for a car to travel from the town school to the county seat is t t = s 6v, and the time for teachers and students is 10 60 = 1 6

The time it took for teachers and students to walk and encounter a car that had been repaired was t

The distance traveled during this time is TV

The rest of the journey takes the S-TV 6V

The time shared in this process is:

1 6+t+(stv) 6v=t+1 2=s 6v+1 2 simplification yields t=2 5

Both 24 minutes.

Assuming that it took T minutes for teachers and students to walk to the breakdown of the car, because the speed of the car was 6 times that of walking, but they had to travel twice as long as this distance (to the school and then back to the breakdown), so the car took T 3 minutes, and the teachers and students set off at 7:10 and were half an hour late, so it took 20 minutes more on this section of the road, thus.

t-t/3=20,t=30

Therefore, the time spent on repairing the car on the way is: 30 + 10 = 40 (minutes).

Let the speed of the person be v, the total distance is s, and the delay time is x, then s 6v+30=x+[s-(x-10)v] 6v, which means that the original time plus half an hour is equal to the time of delay plus the time of the car after that, and the time to go after the car is sorted out to 30=x-(x-10) 6

Solution x = 34 minutes.

Hope it helps.

Parking time to school t1 Walking to school parking time t2 is known by the ratio of speed time: t1 = 1 6 t2

Late arrival time 30 = 10 minutes + more time to walk than drive (5 6 * t2).

t2 = 24 repair time t (minutes) = t1 + 10 minutes + t2 = 38

Yes, 38 minutes, the solution process is as follows.

Assuming that the total distance is S, the person walks to D and encounters the car, the speed of the car is V, and it takes T minutes to repair the car, then the time relationship is as follows.

It turns out that the car shares the time t1=2s v from the county seat

From a human point of view, the present time t2=10+s v+6d v+(s-d) v=10+(2s+5d) v

From the perspective of the car, the present time t2=2(s-d) v+tt2-t1=30 =10+5d v=t-2d v can get t=38

Proof: If you connect an OE, then OE AB

Make of cd at point F

ab=cdof=oe

i.e. the distance from the point o to cd is equal to the radius of the small circle.

The cd is tangent to the small circle.

Connecting ao, bo, co, and do, that is, the triangle abog and the triangle cdo congruence, and the perpendicular lines of ab and cd are made by o, which are oe and of, that is, oe is equal to of, so cd is tangent to the small circle.

1 There is no single answer.

6 + 36 3 - 2) *4 - 1 = 639 + 9 + 9 + 9) 9 = 129 + 9 + 9 + 9) 9 = 202 If each boat goes from 3 to 5 people, that is, two more people, the extra 20 people will be able to sit just right. This means that the extra 20 people can be arranged on all the boats in the same way that there are 2 more people in each boat. 20 people, 2 people per boat, just right, what does this mean?

Description of exactly 10 boats. If 10 boats are filled with 5 people each, there are 50 people.

（1）【（6+36÷3）-2】×4-1=63（2）（9 + 9 + 9 ）/ 9 + 9=12（3）9 + 9 + 9 + 9 ）/ 9=202.The number of children must be a multiple of 5.

If there are n boats, the number of children is 5n

5n-20）/3=n

n=10 There are 50 children and 10 boats.

1. Fill in the symbols or parentheses to make the equation true

2. There is a group of children who rent a boat to play, each boat seats three people, there are 20 more people, each boat seats five people, just full, how many boats and children are there?

Typical Yingkun problem, boat (20-0) (5-3) = 10 people: 10x3+20=50

[6+(36÷3)-2]×4-1=63

Set up x ships are listed by the title.

3x+20=5x

The solution is x=10

Because children = 5x = 10 * 5 = 50

A: 10 boats, 50 people.

5 3 8 (edged).

2) What is more than 200, the maximum is a flat angle, 180, what does 200 mean?

is a regular dodecagon, with an inner angle and 12 150 1800 4,

1. (1) The sum of the inner angles of the quadrilateral is 360°, which can be pushed, and it can be composed of 3 quadrilaterals, so there are 12 edges, except for the four sides, which coincide in pairs! The middle line segment is equivalent to the auxiliary line! You can go: the result is 8 sides.

2) You made a mistake: it should be 20°. Let the outer angle be n, then 3n+20°+n=180°, and the outer angle n=40°, then the inner angle is 140°, so the center angle of the opposite side is 40°. The number of edges a=360° 40°=9

2. The internal error angle is equal: c= b=51° so aob=180°- c- b=87°

3. The center angle of the side is equal to 30° and the inner angle is equal to 150°, the number of sides n=360° 30°=12 Launch: The inner angle has 12 inner angles, so the sum of the inner angles is equal to 1800°

4. The conditions seem to be a little problematic, you see!