Math problems, come in and take a look at high school problems

2: It is known that two circles intersect at two points (1,3) and (m,1), and the centers of both circles are on the straight line x-y+c 2=0, so the common chord equation is: y-3=-1(x-1), so x+y-4=0, because (m,1) is on a common chord, m=3;

The midpoint is on the conjoined line, i.e. (2,2) is on the conjoined line, so c=0, so m+c=3;

3: Let the coordinates of c be (x,y), the radius of the circle c is r, the center of the circle x + (y-3) = 1 is a, the circle c and the circle x + (y-3) = 1 are tangent, and the line y=0 is tangent |ca|=r+1,the distance from c to the straight line y=0 d=r

ca|=d+1, that is, the distance of the moving point c fixed point a is equal to the distance to the fixed line y=-1.

From the definition of parabola, the trajectory of c is parabolic

Therefore, choose A<>

1 (x-2)²+y+1)²=1

Let the coordinates of the midpoint be (x1,y1), then the corresponding point on the circle (2x1-4,2y1+2) (i.e., point p is connected to this point to get the midpoint).

Because the point is on a circle.

So there is (2x1-4) +2y1+2) =4, so simplify it to (x-2) +y+1) =1

2 m+c=3

From the properties of the circles, we can know that the intersection of the two circles is symmetrical with respect to the centers of the two circles.

So the midpoint of the two intersections is at x-y+c 2=0

Because the two points are (1,3)(m,1).

So the midpoint is [(m+1) 2 ,2].

Bringing in the linear equation yields m+c=3

3 A is clearly a question

The distance from that moving point to point C (C is the center of the circle C) = the distance from that point to y=0 + 1 (the length of one more radius because it is tangent).

That is, the distance to the fixed point is a fixed value different from the distance to the fixed line.

This is the parabolic judgment mark.

...... by the midpoint equationx=(x+4) 2,y=(y-2) 2, so, x=2x-4, y=2y+2, substituting x +y =4, we get (x-2) +y+1) =1

It is ...... by the tangent midpointThe intersection of two circles is connected perpendicularly to the center of two circles.

From the midpoint, the point (1+m) 2, (3+1) 2 is on the straight line, so you go in and solve the ......

In fact, you use the nature of the parabola, you draw it, connect the centers of two circles, and then make the center of the circle perpendicular to y=0.

You can get a parabola with x + (y-3) = 1 as the focus and y=-1 as the standard.

1 2, because the coefficients of OA and OB add up to 1, and the Hehe vertical is on a line with m, a, and b, because OM = two-thirds OA + one-third OB, so AM = 1 2AB (because the OA coefficient is 2:1 than the OB coefficient, so AM: AB = 1:

2, as if it is a hunger theorem) hope

Take the midpoint of ab e, then ve ab, because the plane vab plane abcd, ab is the intersection line, so ve plane abcd, as ef ac, can prove ef ac, so the angle vfe is the angle sought, let ab = a, then ve 3a 2, ef bo 2 a (2 2).

The tangent of the dihedral angle v-ac-b is ve ef 6

1.p (no less than 2 people each) = p (3 miss, 2 translators) + p (2 miss, 3 translators) = [c(5,3)c(4,2)+c(5,2)c(4,3)] c(9,5).

10x6+10x4)/126

2.3x2x2=12 species.

3. f'(x)=3x2-3b=0 x=+ - b minimum, then f''(x)=6x>0, x>0, x= b, then 0< b<1 --0y'=1/x

Lingy'=k (i.e., find the tangent of the slope of lnx with k) to get x=1 k, substituting y=lnx, the tangent is (1 k, -lnk) the tangent is also on y=kx, substituting -lnk=1, then k=e (-1)=1 e selects c

There are many similarities between the two series, equal difference and equal ratio, in fact, the proportional series is a difference series after taking the logarithm. It is helpful to keep the following points in mind:

1.Both are two unknowns, the first term A1 and the tolerance (ratio) q, need two conditional columns and two equations to solve.

is the difference (quotient) of two adjacent terms

3.Either term is the arithmetic (geometric) average of the two terms:

That is, the equal difference series: 2an=(an-1)+a(n+1), and the proportional series an 2=a(n-1)a(n+1).

4.Sum by the first term and the tolerance (ratio): na1+n(n-1)q 2; a1[1-q^(n-1)]/(1-q)

5.Sum by the first and last terms: (a1+an)n 2, a1[1-q (n-1)] (1-q), q=(an a1) [1 (n-1)].

6.The odd term is summed by n times (to the power) of the intermediate term am: nam, (am) n

7.Even terms are summed by n 2 times (power) of the middle terms am, am+1 and (product): (am+am+1)n 2, (amam+1) (n 2).

You can use whatever the topic asks you for!

Once the geometrical features are grasped, the problem is solved.

The solution is as follows:

What about my answer?

This is how the value of parameter a will be.

Seek out, the process sees**.