AB is the diameter of the circle O, C is a moving point on the semicircle, CD ⊥ AB connects Co, CP bisects ∠ OCD, does the position of P change with the position of C?

AB is the diameter of the circle O, C is a moving point on the semicircle, CD ⊥ AB connects Co, CP bisects ∠ OCD, does the position of P change with the position of C?

Point P does not change
Prove: connect OP DP, because OP = OC, so angle OCP = angle OPC, and angle OCP = angle DCP
So angle OPC = angle DCP
So OP parallel CD
And CD is perpendicular to AB, so OP is perpendicular to ab
And P is on the circle, so there is always OP perpendicular to AB, P is a fixed point

If the solution of the equation X-1 / 2x + a = 1 of X is positive, then the value range of a is

2x+a=x-1
2x-x=-1-a
x=-1-a
The solution of this equation is positive
∴-1-a＞0
-a＞1
a＜-1

The hyperbola passing through the origin has a focal point F (4,0), 2A = 2. Find the trajectory equation of the hyperbola center

Let the hyperbola center coordinate (x, y) be exactly the midpoint of two focal points, one focal point is (4,0), then the other focal point coordinate is (2x-4,2y). The origin (0,0) is a point on the hyperbola, and the distance difference between the two focal points is the fixed value 2A = 2, so | √ [(2X-4) & sup2; + (2Y) & sup2;] - 4 | = 2, the solution of hyperbola center line trajectory is obtained

It is known that X1 and X2 are two parts of the equation 3x square 2x-5 = 0. Solve the equation and find the absolute value of x1-x2
X1 and X2 are two incomprehensible equations of equation 3x ^ 2-2x-5 = 0. Find the absolute value of x1-x2

X1 and X2 are two parts of the equation 3x ^ 2-2x-5 = 0
The results are as follows
x1+x2=2/3,x1x2=-5/3
because
(x1-x2)^2
=(x1+x2)^2-4x1x2
=(2/3)^2-4*(-5/3)
=64/9
therefore
|X1-x2 | = radical (x1-x2) ^ 2 = 8 / 3

E. F are the middle points of edges A1A and C1C of cuboid a1.b1.c1.d1-abcd, respectively

Let G be the midpoint of BB1. Let b1gcf be a parallelogram. B1F ‖ = GC
Ge ∥ BA ∥ CD, ∥ CDEG is parallelogram, GC ∥ ed. ∥ B1F ∥ ed
The b1edf is a parallelogram

The square of a few is eight

If the square of a few is equal to 8, it is to find the square root of 8
The square root of 8 is: √ 8 and - √ 8, and √ 8 = √ (2 × 2 × 2) = √ (2 & sup2; × 2) = 2 √ 2
So the square roots of 8 are 2 √ 2 and - 2 √ 2
So the square of 2 √ 2 and - 2 √ 2 is 8

Is the square product B of 3A + 2b-c a polynomial?

Is the square of 3A + 2b-c multiplied by B a quadratic trinomial? Why?

If a + B = 6, A-B =?

a+b=6
Then (a + b) ^ 2 = a ^ 2 + 2Ab + B ^ 2 = 36
Because a ^ 2 + B ^ 2 = 18
So 2Ab = 36-18 = 18
（a-b)^2=a^2+b^2-2ab=18-18=0
So A-B = 0

The two adjacent sides of a parallelogram are 4cm and 6cm respectively. The distance between a group of opposite sides is 4.5cm. The area of this parallelogram is______ Square centimeter

4 × 4.5 = 18 (square centimeter); answer: the area of this parallelogram is 18 square centimeter

Solution equation: X / (x-1) = 1 + 3 / (x ^ 2 + X-2)

x/(x-1)=1+3/(x^2+x-2)
Multiply both sides of the equation by X & sup2; + X-2 at the same time to get
Then x (x + 2) = x & sup2; + X-2 + 3
Then x & sup2; + 2x = x & sup2; + X + 1
Then x = 1
Test: substitute x = 1 into the original equation, denominator = 0, so x = 1 is an increasing root
So the original equation has no solution