A high school math problem to solve, a high school math problem to ask for a detailed solution

The center of gravity G of the triangle ABC

g[(x1+x2+x3) 3,(y1+y2+y3) 3] analysis: Let the midpoint of ab be d

So d abscissa 2, and the center of gravity theorem tells us ad=3gd, so x3- 2=3 2}, giving x= 3

The same goes for the ordinate.

The midpoint of BC D((X2+X3) 2, (Y2+Y3) 2)G is on AD, and AG=2Gd

So (xg-xa,yg-ya)=2(xd-xg,yd-yg)1)xg-xa=2xd-2xg,3xg=xa+2xd=x1+x2+x3,2)yg-ya=2yd-2yg,3yg=ya+2yd=y1+y2+y3,so g( (x1+x2+x3) 3,(y1+y2+y3) 3 )

In the triangle ABC.

a(x,y) b(p,q) c(j,k)

Center of gravity abscissa = (x+p+j) 3

Center of gravity ordinate = (y+q+k) 3

Vector EC · vector EM = vector (EB + BC) · vector (EB + BM).

Vector EB + Vector EB Vector BM + Vector BC Vector EB + Vector BC Vector BM

eb|²＋0+0+1*，

Select [C] for this question

If it is known that the circle is a unit circle, the trajectory is a circle; When it is known that the circle is not a unit circle, the trajectory at this point is an ellipse.

a is on a circle with the center of the circle at the origin.

So|oa|=r (radius).

ob|=1/|oa|=1 r is a fixed value.

Then the trajectory of b is also a circle selection a

a。It must be round.

If OA 1, then OB 1, the two circles coincide;

If OA 1, then OB 1, the second circle is inside the first circle;

If OA 1, then ob 1 and the second circle is outside the first circle.

You should choose [C] for this question

If the circle is known to be a unit circle, the trajectory is a circle; If it is known that the circle is not a unit circle, the trajectory is an ellipse.

Because f(x-1) and f(x+1) are odd functions, so f(-x-1) = f(x-1) and f(-x+1) = f(x+1).

f(-x+1)=-f(x-3).

Simultaneous f(x-3)=f(x+1) i.e. the function f(x) is an odd function with period 4.

So f(x+3) is as odd as f(x-1).

So the reward doesn't have to be so high Everyone has always liked these questions and is happy to learn.

A multiply the first two and end up with 1 2cos2x