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Because the straight line where its bottom edge is located passes through the point (3,-8), the linear line equation of the bottom edge CB is Y+8=K(X-3), and the angle between it and the straight line where the waist AB and AC is located is 7X-Y-9=0 and X+Y-7=0 is equal to find K.

Supplementary method: The angles are equal, so the absolute value of k-7 is 1-(7)k=the absolute value of k+1 is 1-(-1)k

Solve k. where 7 and -1 are the slopes of the two straight lines of ab and ac.

7x-y-9=0

x+y-7=0

The coordinates of the vertices obtained by the biequality on the synthesis are (2,5).

The slope k of the angular bisector of the angle bac can be found by passing through the vertex a, and since the angular bisector is perpendicular to bc, the slope of bc can be found by -1 k, and the equation of cb can be obtained from the point oblique.

Let the abscissa of point b be a and the abscissa of point c be b, then the coordinates of point b are (a,7a-9), the coordinates of point c are (b,7-b), and the coordinates of point a are the solutions of the two equations (2,5).

abc is an isosceles triangle, ab=ac, so (a-2) 2+(7a-9-5) 2=(b-2) 2+(7-b-5) 2

a-2)^2+(7a-14)^2=(b-2)^2+(2-b)^2

50(a-2)^2=2(b-2)^2

b-2=±5(a-2)

BC passes the point (3,-8), so (7a-9+8) (a-3)=(7-b+8) (b-3)=>b=(9a-12) (2a-1) is substituted into the above equation a=2 or 2a-1= 1

A=2, B=2 (ABC is the same point and therefore not true) or A=1, B=-3 (the point is not on BC, on the BC extension line) or A=0, B=12

Therefore, b is (0,-9) and c is (12,-5).

The linear equation for cb is y=x 3-9

Is the old university vectored?

It's relatively simple to do it with vectors.

Because the slope of the equation of the straight line of the waist is 7,-1

Lian Li finds the vertex as (2,5), and then passes the point (3,-8), and makes a sketch judgment, BC is at the bottom right of the vertex (the judgment is to set the direction vector).

So the direction vectors of the two lines are (-1, -7), (1, -1) [in the same direction as the vectors ab and ac, respectively].

The direction vector of the bottom is (1,k).

It is because the angle between the bottom and the two waists is equal.

So (1,7)(1,k) |(1,7)||1,k)|=-(1,-1)(1,k)/|(1,-1)||1,k)|(The angles between the two vectors and the direction vectors of the straight line are complementary, and the cosine is the opposite of each other).

solution, k = 1 3, over (3, -8).

So the linear equation y=x 3-9

This is method 1, which is characterized by being difficult to think and prone to error, but it is easy to calculate.

Method 2 (this is difficult to calculate, just talk about the idea).

Let the slope ky+8=k(x-3).

Separate him and the two waists to untie the bottom ends.

Then find the bottom edge of the vertical line.

Then connect the two waists to find the apex.

The vertex is on the middle perpendicular line, which in turn solves k

There is no easy way to segment functions, segment-by-segment analysis.

Profit s=n(x)*p(x)-4n(x)=write out in segments and analyze them one by one.

The mobile phone is hand-made to watch it.

In the second step, sin cos is raised as a whole3

The coefficients become 1/2 and 2/2 root 3, respectively

The meter can be written as sin30 and cos30 respectively

Then the sinusoidal formula sin (a plus b look).

Conversion yields the result.

The distance from the midpoint of the string AB to the alignment = (the distance from point A to the alignment + the distance from point B to the alignment) 2 = (af + bf) 2

ab 2 The next step should be that it is synchronous, using the relation of vectors, to solve the straight line y=k(x-1).

a(x1,y1) b(x2,y2)

x1-1) (1-x2)=3 solution k

ab uses the chord length formula.

It is known that the circle c: x 2+y 2+2x-4y+3=0. A tangent line is drawn from the point p outside the circle c, the tangent point is m, o is the coordinate origin, and there is |pm|=|po|, begging for a messenger|pm|The coordinates of the smallest point p.

Solution: According to the topic, satisfied|pm|=|po|, and make |pm|The smallest point p, at least a little on the oc.

Let p(x,y).

Circle c: x 2+y 2+2x-4y+3=0==>(x+1) 2+(y-2) 2=2

oc equation y=-2x

Set |pm|=|po|=m

pm|^2+2=(|oc|-|po|)^2

i.e. m 2 + 2 = ( 5-m) 2, the solution is m = 3 5 10

then x 2 + y 2 = 9 20 ==> 5x 2 = 9 20 ==> x1 = -3 10, x2 = 3 10 (round).

then p(-3 10,3 5).

A point outside the circle, there are two tangents to the circle, and the distance from one point to the two tangent points is equal, to make this distance equal to po, then the two tangent points and the origin o must be co-circular, and the position of the point p is different, so that the radius of this circle is different, which one has the smallest radius? It must be the radius of the circle in which the tangent point and the origin point form an equilateral triangle, thus satisfying the |pm|=|po|, and make |pm|The smallest point p, not at least, should be necessarily on the oc.

Let pm=po=x

pc=sqrt(x^2+2)

co=sqrt(5)

PC+PO>=CO (triangle).

sqrt(x 2+2)+x>=sqrt(5)xshift, and then squared.

x 2+2>=x 2-2x times the root number 5+5

x>=3 10 times the root number 5

Take the minimum value of p on co.

The coordinates are (3 10, 3 5).

PC+PO>=CO is equal to the equation, which means that the point p is on CO, and P is not on CO, and PC+PO>CO is a triangle

He has a very clear idea, and it can be seen that his mathematical skills are still very good, so I will briefly talk about it.

The second condition is |pm|=|po|, which can be equivalent to the two vectors perpendicular to om and pn (n is the midpoint of om). This should simplify a lot of calculations.

The method is a bit troublesome, and I didn't expect a good way, but I don't think it's right for you to solve the point p in the end.

It should be a perpendicular line made by the dot O to the P-point trajectory, and the point to the straight line is used, and the OC is not necessarily perpendicular to the P-point trajectory.

It's not a complicated question.

The list is: h tan29 - h tan38 = 200

By looking up the calculator we can get h=

Here we should pay attention to the relationship between right triangles h bottom edge = tan diagonal If you have any questions, you can continue to ask me.

If the distance from the second observation point to the foot of the mountain is x, the distance from the first observation point to the bottom of the mountain is x+200

tan29=h/(x+200)

tan38=h/x

Solving the above system of binary linear equations yields h

f(1)=10 know a+b+c+d=9

f(2)=20 8a+4b+2c+d=4f(3)=30 know27a+9b+3c+d=-51f(10)+f(-6)=11296+784a+136b+4c+2dling784a+136b+4c+2d=x*(a+b+c+d)+y*(8a+4b+2c+d)+z*(27a+9b+3c+d).

The system of equations is x+8y+27z=784

x+4y+9z=136

x+2y+3z=4

x+y+z=2

The solution is x=64, y=-126, z=64, and if the equation is satisfied, then the solution of the system of equations is x=64, y=-126, and z=64 then 784a+136b+4c+2d=64*9-126*4-51*64=-3192

then f(10)+f(-6)=11296-3192=8104

f(10)+f(-6)=11296+784a+136b+4c+2d (1)

f(1)=10 yields: a+b+c+d=9

f(2)=20 gets:8a+4b+2c+d=4, that is:d=4-8a-4b-2c (2)f(3)=30 get:

27a+9b+3c+d=-51 simplify and eliminate c, after d: -50=12a+2b (3) substitute (2) into (1) to obtain: 11296 + 768a + 128b + 8 (4) to substitute (3) into (4) to obtain:

Question 9, using the value of x, first prove that x 2-1=-x, so the original formula = x 4-2x 2+1+3x 2+2x-2=(x 2-1) 2+3x 2+2x-2=4x 2+2x-2=4(x 2-1)+2x+2=-4x+2x+2=-2x+2, and then it will not be counted.

In question 10, you should calculate (x-1)*(x-4) and (x-2)*(x-3) first.

Question 11, you just have to do it head-on.

Your original formula =-x 4+[(y+z) 2+(y-z) 2]x 2-[(y+z)(y-z)] 2

I don't have to forget about it then.

1) Define the domain as r

a>0y=a+1 a>=2 root number (a*1 a) = 2 value range [2,+

2) This is a double-hook function, so odd functions.

3) Let a=2 x

then a>0'

y=a+1/a

Then 01, increase the function.

a=2 x is the increment function.

So 0<2 x<1, subtract the function.

2 x>1, increment function.

So increase the interval (- 0).

Subtract the interval (0,+.)