Middle School Mathematics Process Thank you, Middle School Mathematics want the process, thank you

This answer should be two One is 1 and the other is 5 under the root number and 1 first: the radius is 1 and the chord ab is the root number 2 It can be seen that the angle AOB is 90 degrees, so the angle bao is 45 degrees, so the angle pab is 45 degrees, which exactly constitutes the isosceles right triangle pab, so we can get that pb is 1

Let's talk about root 5:

Let the last point b be b1 and this time b bb1 is the diameter of the circle, because the last time pb1 is 1, the diameter is 2 and the angle pb1b is 90 degrees, so pb is the root 5

Hopefully.

1 and root number 5,1) When the points A and B are on either side of the op, the angle AOB is determined to be 90 degrees according to the length of OA, OB and AB, and then it can be proved that the quadrilateral OAPB is a square and PB1 is equal to 1

2) When point A and point B2 are on the same side of OP, because the angle AOB1 is 90 degrees, the same angle AOB2 is also 90 degrees, B1B2 is the diameter of the circle O, the diameter is 2, PB1 is 1, and the angle PB1B2 is 90 degrees, so PB2 is the root number 5

No way! The radius of the circle O is 1, ab is the chord of the circle O, how can ab be longer than the radius???

The radius is 1 and the diameter is 2 and the largest chord in the circle is the diameter.

ab=2√2>2 ？？Is it a string inside a circle?

！From the meaning of the title, the coordinates of the intersection point p1 (2, 2) p2 (root 2, 2 root 2) are known, so the coordinates of other points are known.

Answer: 8 6 root 2

2) Let the coordinates of point A be (x,y), so s=-1 2xy=1, i.e., xy=-2, then y=-2 x

and a is on y=m x, so m=-2

So y=-x+1

The intersection point a of the two sub-formulas is (-1,2).

a-1) b-(b-1) envy bend a +1=0 pass score.

a(a-1)+b(b-1)+ab=0

a2+b2+ab=a+b

i.e. (a+b)2—(a+b)=ab

By the fundamental inequality: ab (a+b)2 4

That is, (a+b)2—(A+B) (a+b) (a+b) Solve the unary quadratic inequality as a whole by reading 2 4 and get 1

1) x=0, no matter what value m takes, there is always y=1, that is, it passes through the fixed point (0,1).

2) Because there is an intersection point, i.e., the equation mx -6x+1=0, the only solution, so 1m=0, which is clearly true;

2. m≠0

6²-4m×1=0

4m=36m=9

Proof: The quadrilateral ABCD is a diamond.

bc＝dc，∠bad＝∠bcd＝α

The line segment CE rotates clockwise around point C to get CF

ce＝cf，∠ecf＝α，bcd＝∠ecf∴∠bce＝∠dcf

In BEC and DFC, BC DC, BCE DCF, CE CF BEC DFC(SAS), BE DF.

The triangle DCE is equal to the triangle ACB, so the angle A = angle C, and because the top angle is equal, that angle is also a right angle.

Answer: Filial piety omitted to clear the case, and it was searched and destroyed before the incident.

<> steps are as follows.

Connecting the state circle AD, because it is an isosceles triangle, so