A middle school math that requires a detailed solution process.

Let ac=2x, then cd=3x, db=4x; Then there is ab=9x.

And p is the midpoint, so ap=

PD = cm can be pushed out x = 4 cm to get AB = 36 cm, CD = 12 cm;

pd:pc=:5

FYI!

ac：cd：db＝2：3：4。So ab 2+3+4 9

ap＝2/9 pd=ad-ap pc=ad-ac

That's the general idea. I hope you think about it for yourself, and you will definitely be able to do it.

Match at a to 99 when the Kaysen and for 100. At this time s99 199. Then, a to 100 Kaysen and is:

s1+s2++s99+s100）/100=(9900+s100)/100=（9900＋s100）/100＝（

Solution: Let Xiao Ming's velocity be V1, Xiao Ying's velocity be V2, and AB's distance between the two places is S.

According to the title, (v1+v2)a=s,,(v1—v2)b=s, so (v1+v2)a=(v1—v2)b

v2a+v2b=v1b—v1a

Therefore, v1 v2 = (a+b) (b—a).

You may wish to set Xiao Ming's speed to s1, Xiao Ying's to s2, a(s1+s2)=b(s1-s2).

s1/s2=（a+b）/（b-a）

Xiao Ying's velocity is x, Xiao Ming's velocity is y, and the distance between a and b is l(x+y)*a=l

yb-xb=l

y=(a+b)/(b-a)x

1. (20 3) A to the power 8 15) B to the power (9 16) C = 4

It can be reduced to: 20) A to the power (3) A to the power 8) B to the power (15) B to the power 9) C to the power (16) C to the power = 2 squared.

2) 2 A (5) A (3) A (2) 3b [(3) B (5) B] 3) 2C (2) 4C.

3 [(2c-a-b) power] 2 [(3b-4c+2a) power] 5 [(a-b) power] = 2 squared.

So: 2c-a-b=0

3b-4c+2a=2

a-b=0 gives a=b=c=2

2, (3n-3m power of a, 33m power of 2011).

It can be reduced to.

a to the nth power a to the m power) to the 33rd power - 33) to the 2011 power.

8-33) to the power of 2011.

-25) to the power of 2011.

Yes, question 2 a*3n-a*3m-33=(a*n)*3-(a*m)*3-33=415

So in the end, it is equal to 415*2011 Just leave it as it is, and junior high school will not ask for such a large number.

Make point a with respect to the x-axis of the point of symmetry a'(-8,-3), point b'(4,5) of symmetry with respect to the y-axis

Link A'b', and the x and y axes have intersection points d and c respectively, and the circumference is the shortest.

Using the method of undetermined coefficients, a is obtained'b'The equation for a straight line y=3 4x+2.

Then find the coordinates of the intersection points c and d. c(0,2)d(-8 3,0) is known to find the coordinates of the intersection of the two axes. Let x=0, y=0 respectively) i.e. m=-8 3, n=2

So: m n=-4 3

The score line is inconvenient to play, and the steps are a little messy. But the thinking is quite clear. Hope.

This question is a typical shortest distance problem. It is a more difficult type for the comprehensive use of the middle school axis to Wu Xiang and the plane Cartesian coordinate system. It is very similar to the shortest distance for a sheep to go to the river to drink, then graze in the meadow and back again.

Draw a coordinate plot and list an equation based on the graph! Hands-on work will cultivate interest.

Root number + root number + root number + root number = perimeter.

The perimeter of the minimum clear state is obtained, and m,n is obtained

First of all, the distance of ab is fixed, and c, d move on the x and y axes. Make point a0(-8,-3) with point A with respect to the x-axis, and point B0(4,5) with point b with respect to the y-axisFrom the nature of symmetry, we can know that AD=A0D, CB=CB0, because the distance between the two points is the shortest, so A0B0 is greater than or equal to A0D, DC, the sum of CB0, and C D is the intersection of A0B0 and the XY axis of the roll.

Circle o outcut regular hexagon side length.

a=2/√(3)*r=2/√(3)

s = area of the circle o inscribed regular hexagon - area of the circle o.

6*√(3)/4*a^2-π

The area of this part is divided into 2 parts, that is, the upper part of the 6 corners and the circular part in the middle, and the area of the 6 corner parts is (1 3-6)*6

The middle circle part (316 3-80 3), so the total area is 6 3+ (313-80 3).

The radius of the circle is 1, then the diameter is 2, in fact, the place where the circle is not is the four corners of the square, and the same is the square with a side length of 2, the area of the square is 4 square centimeters, and the area of the circle is, so, the area of less than is .