A comprehensive question for the third year of junior high school, and a question for the third year

1) y=(3 of the root number of 3)*(x-1)*(x-7 2) abc is an isosceles triangle, if qab is similar to cab, then the straight line x=4 has and only intersections with the parabola c and has no solution.

If ABC is similar to CBA, then Q is the intersection of the straight AC and the parabola in the X 0 region. The straight line ac and the parabola have only two intersections of a c, and there is no solution in the region of x<=0.

So the QAB does not exist.

1).Let y=a(x-4) 2-3 (a is not equal to 0) y=ax 2-8ax+16a- 3

ab=6 b 2-4ac) a=6 i.e. (8a) 2-4a(16a- 3) = (6a) 2

a= 3 9 can be substituted.

2).Let q(x,y).

bc=2√3

ab/bc=qb/ab

36=2√3√[(x-4)^2+y^2]①y=√3/9(x-4)^2-√3②

Simultaneous solution x,y is sufficient.

If it cannot be solved, it proves that it does not exist.

The first question is y=(3 1 2) 9*(x-4) 2 -3 1 2 and the second question. Let it exist. Dot: Q

Let's first take the **triangle ABC, it is not difficult to find that AC=BC, the angle BAC=30 degrees, since the parabola is symmetrical about X=4, you might as well set Q on the right end of the parabola, then the angle AQB=30 degrees, AB=BQ=6.

We can find the equation of the straight line where aq is located, substitute it into the parabolic equation, find the coordinates of the q point, and then find bq from the two-point distance formula to see if it is equal to 6

With y1-y2, we get y=--3 8)x+36-(1 8)xsquared 15 8x+59 2

1/8x^+12\8x+13/2

1/8(x-6)^+6

It can be seen that when x is equal to 6, the profit is the largest, and it can be seen from the title that before May Day, the closest to 6 is April, and when x = 4, y =

The idea is right.

0 Heng was founded, apparently with two roots.

It is also correct to start writing greater than or equal to, because it is not said that the two are different.

Since it is said that there are two roots, it should be "greater than zero", not "greater than or equal to zero".

Because the original equation has two integer solutions.

So 8m+4 must be a square number and a multiple of 4.

I don't see this point of yours.

But I have a better suggestion, which is to use the sum of the two roots and the product of the two roots, if I am not mistaken, the sum of the two should be equal to 2 (m+1), and the product of the two roots is equal to m, and then we will discuss it. But I've been graduating from high school for n years, and I can't remember the relationship, so please check it yourself!

1.It should be relatively easy to prove.

Four-point circle (this should have been learned, right?) ), so let AD and CE intersect at point F, and the triangle EFD is similar to the triangle AFC, so de ac=ef af=1 root number 2

(1): The two corners correspond to proportions, and the two triangles are similar. （2）

The side length of the small square ad=root(A2+B2)=root(20+30)=root(50)=5,root(2).

Small square area = AD 2 = 50

The radius of the small circle = ad 4 = 5 root number (2) 4

The area of a small circle = *(5 root number(2) 4) 2=25* 8The area of the yellow flower = the area of the four small circles 25* 8*4=25* 2The area of the green grass = the area of the small square - the area of the yellow flower = 50-25* 2

First, press the surface of the water and draw the symmetry point p of the airship' ，pp'The intersection point with the extension of the horizontal line where A is located is set to O.

Then: Angular p'AO=60°, angle PAO=45°So, PO=OA, P'0 = root number 3 * oa then: (op.)'+PO) 2 = 50 + PO yields: OP=100 (root number 3 - 1).

1.From the title, there is y=x(50-2x)=-2x 2+50x (0 max. Individuals give two methods Junior high school suggests the first one.

1) With f(x)=a(x-h) 2=k

y=-2x^2+50x

2(x^2-25x)

2(x^2-25x+

2 (So, when x=, y has a maximum.)

2) Derivative. f'(x)=-4x+50

When f'(x)=0, that is, when x=, the original function obtains an extreme value. Since the original function is convex, the extreme value is the maximum.

Substituting x= into the original function gives y=

Happy learning.