High school math questions 20 questions to detail the process, thank you

2x²+(6-2m)x+m²-4m+3=0。。。There is a unique solution.

Therefore, δ=(6-2m) -4*2*(m-4m+3)=0m-2m-3=0....The solution is m=-1 or 3

Therefore, the tangent equation is x+y+1=0 or x+y-3=02) from the problem, it can be seen that PMC is a right triangle, and the angle PMC is a right angle (C is the center of the circle) pm|²=|pc|²-mc|²。where mc is the radius.

x1+1)²+y1-2)²-2...Due to typing problems, x is used for x1 and y for y1

x²+y²+2x-4y+3

po|²=x²+y²

By |pm|=|po|=>|pm|²=|po|²x²+y²+2x-4y+3=x²+y²

From the quadratic function characteristics, we can see that when x=-3 10, the above equation takes the minimum value, and y=3 5 so the point p(-3 10,3 5).

1) Rounding is the standard formula, (x+1) 2+(y-2) 2=2, the center of the circle is (-1,2), according to the meaning of the problem, the tangent equation is x+y+c=0, the tangent is that the distance from the center of the circle to the straight line is equal to the radius, and the column equation calculates c=1

I didn't understand the second question.,I'm sorry.。。。

This problem requires experience: the parabola diameter is 2p long

Jiao Ji points to the distance from the line p, and takes the diameter of the diameter to make the garden tangent to the line, indicating that the garden is what is sought, and ab is the diameter, and the length is 1 2

I'm a sophomore now, but I still like math problems like this, and I've seen countless rounds of difficult problems, and I miss them so much.

1.The inner, outer and center of gravity of a regular triangle are one.

1) It is easy to obtain the radius of the inscribed circle and the radius of the circumscribed circle, and form a 30-degree triangle with the radius of the inscribed circle being the short side of the right-angled side and the radius of the circumscribed circle being the hypotenuse of the right-angled side. Therefore the radius of the inscribed circle: the radius of the circumscribed circle = 1:

2 (2) due to the radius of the inscribed circle: the radius of the circumscribed circle = 1:2

Therefore, the radius of the inscribed circle: height = radius of the inscribed circle: (radius of the inscribed circle + radius of the circumscribed circle) = 1:

3 So the centroid distance: radius: height = inscribed circle radius:

Circumscribed circle radius: height = 1:2:

32.Let the radius of the circle be r

Then the side length of the circumscribed regular hexagon = 2 (3) r 3, the height of the circumscribed regular hexagon = r then the side length of the inner regular hexagon = r, the height of the inner regular hexagon = (3) r 2 the ratio of the side length of the inner regular hexagon to the outer inscribed regular hexagon = r: 2 ( 3) r 3 = ( 3): 2 The height ratio of the inner regular hexagon to the outer inscribed regular hexagon = (3) r 2:

r=(√3）:23.Let the radius of the circle be r

then the inscribed n-sided side length a=r*sin( n) then the circumscribed regular n-sided side length b=r*tan( n)b=a cos( n).

4.The radius of the circle is r

then the inscribed regular n-sided side length an=r*sin( n) then the inscribed regular 2n grilateral side length a2n=r*sin( 2n) then the inscribed regular 2n engonal height b2n=r*cos( 2n) inscribed regular 2n engonal area s2n=a2n*b2n*2n 2=(n 2)*r*r*sin( n)=n*an*r 2

s8=4*a4*r/2=2*(√2）*r*r

When x 0, f(x)=-x, it is easy to know f(x) 0.

Let f(x)=t, then f[f(x)]=f(t)=(t-1 t) 6

You can use the binomial theorem to solve the problem of constant terms. The known constant term is c63*t *(1 t).

Is this a fill-in-the-blank question or a calculation question? If it's just a fill-in-the-blank or multiple-choice question, I'll just talk about my difficult thinking, which will help you make these kinds of questions faster without spending a lot of time calculating.

First determine the range of x, greater than 1, it must be a positive number, and then look at this function separately, first of all, the square of 1 x, when x increases from 1 to 6, the function value decreases monotonically, and the function of -6x is also monotonically decreasing, and the constant 6 behind it is left alone, so the function as a whole decreases with the increase from 1 to 6, so the function takes the maximum value when x=1, and takes the minimum value of 6, and the orange value range is greater than -29 and 35 36

Less than or equal to 1

y'=-1 x 3-6<0 x belongs to [1,6), so the function y=1 x squared -6x+6 is a subtractive function when x belongs to [1,6).

Substitute the two endpoint values into the cherry orange spine respectively.

The value range is (1 36-30,1].

m is the upper point of the ellipse x 2 64 + y 2 48 = 1, f1 and f2 are the left and right focal points respectively, satisfying mf1 = 3mf2, what are the coordinates of the m point?

In the problem, we know that a=8, b=4 3, c=4, then f1(-4,0), f2(4,0), let the m point be (x,y). Yes.

x+4) +y =9((x-4) +y), and there is x 2 64+y 2 48=1, and the joint solution equation has x=8 and y=0

I'll tell you the idea.

The process of solving the problem is difficult to explain with a computer, especially if the picture cannot be drawn.

Question 1: First make two high lines, connect the intersection point and another point, and prove that the line is perpendicular to the other side through the relationship of angles, that is, the high line.

Question 2: First make two middle lines, the intersection point and the vertex line intersect on the other side, make a median line of the two known middle lines, and the other one can be obtained through the relationship between the area, and in the same way, the relationship between the area can also be deduced to be a 3rd equal point.

Question 3: The center of gravity is the intersection of the midline, and the heart is the intersection of the bisector of the angle, I probably haven't done high school geometry problems for a long time, and I can't think of a good way. This problem can be solved by using the method of coordinates, establishing a coordinate system to find the coordinates and calculate the length.

Solution: Let t=3-ax, then y=log a[3-ax] is deformed into y=log a(t), and when 01, t is a decreasing function of x, and y is an increasing function of t.

So y=log a[3-ax] is a subtractive function in the defined domain, so 3-ax>0, and a>1, x belongs to the interval.

0,1], so 1 pro: Iridiumian.

g(x)=ax 2-2x+2>0 is true for any x [1,4], and since a=0 is obviously not eligible, the following discussion is based on a≠0, and the final answer is to remove a=0.

There are three situations in the figure below:

They are available in turn:

Blue: 1) The axis of symmetry is to the left of 1, i.e. 1 a<12) g(1)>0 red: discriminant <0

Green: 1) The axis of symmetry is to the right of 4, i.e. 1 a>42) g(4)>0 can be obtained by taking the union of the three.

I'm really sorry, you can't see the ** you send.