Mathematics in the second grade of junior high school, in a hurry, in a hurry

Solution: In ADC and ABE, ADC and ABE are equilateral triangles.

Thus ad=ab ac=ae dab= eac=60 degrees.

and dac= dab+ bac=60 degrees + bac eab= eac+ bac=60 degrees + eac

Thus dac= eab

then ADC ABE (Corner Edge).

Thus be=dc

Connecting ao is obtained by adc abe adc= abe so that a, d, b, o four points are concircular [the circumferential angles of the same arc are equal, then these four points are concircled] then bod= bad=60 degrees.

Thus boc=180 degrees- bod=180 degrees-60 degrees=120 degrees.

1) Prove: Because it is an equilateral triangle, AB=AD (equilateral triangle property), AC=AE (equilateral triangle property), angle DAC=angle BAE (both angles BAC+60°), triangle ADC is all equal to triangle ABE (method: corner edge).

So: be=dc

2) Can't see clearly.

Because they are both equilateral, ab=ad ae=ac, and the angle dac = the angle bae, so the triangle diac is all equal to the triangle bae. So: be=dc

Your figure is vertical, and the neck hurts when you look at it, beg. Immediately after that, I asked a second question.

(1)(a-b) 2=a 2+b 2-2ab=(a+b) 2-4ab=5 2-4*3=13 So a-b = positive and negative root number 13

2) Since x 2-5x+1=0, x is not equal to 0, so both sides are divided by x to get x+1 x=5, so x 2+x -2=(x+1 x) 2- 2=5 2- 2=23

3) A 2+b 2+c 2-4a+4b-bc=(a-2) 2+(b+2) 2+(c-3) 2-17 (This question may be wrong, it should be 9c, -17).

I can't stand it.

1 a-b = root number 13

2 I can't see clearly.

3 Classes tomorrow I'm also in a hurry.

a^2+b^2+c^2-4a+4b-bc≤17(a-2)^2+(b+2)^2+(c-3)^2<=0a=2,b=2,c=3

a+b+c=7

First question, I want to think about it.

A plus b) square is 25, can be a square + b square + 2AB = 252AB is 6, a-b is first replaced by (a-b) square, that is, a square + b square - 2aba square + b square = 25-6 = 19, a-b) square is 19-6 = 13, a-b is the root number 13

That's what it means.

Solution: (1) The diagram shows the relationship between the distance s (km) between them and place A and the time t (hour) traveled, according to the image: A catches up with B in 2 hours;

2) In the figure, L1 passes through the origin and (2,6), so the function relation representing A is: S=3T, L2 passes through (0,3),(2,6), and the function relation is: S=KT+B, and the coordinates of the two points are substituted to obtain:

k=,b=3, so the functional relation of B is: s=;

3) Subtract B's analytic formula from A's analytical, and its value is equal to 3, which can be obtained, and t=4 can be obtained

This is a philosophical question, and Einstein has to concede three points when he comes. It is difficult to answer such a profound question by being contemptible and mediocre.

Let the velocity of a be v, and there is a functional relationship to obtain s=vt, and the velocity of b with an image is v2, then s=3+v2t is obtained when t=2.

v2= v=s/t

s/t-(s/

Are you sure your diagram is correct?

Fold the rectangular piece of paper with length and width as shown in the figure, and cut a square with a side length equal to the width of the rectangle (called the first operation); Fold the remaining rectangle as shown in the picture and cut a square with sides equal to the width of the rectangle at this time (called the second operation).

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The angle dce=60 degrees is obtained from the equilateral triangle, and from the ce=cd, the angle e=30 degrees is obtained by using the outer angle.

Then there is the condition of the middle line, the combination of the three lines gives the angle DBC=30 degrees, then the angle e=angle DBC, so DB=DE