A and B are coprime numbers, their greatest common factor is (), their least common multiple is (). Explain the reason

A and B are coprime numbers, their greatest common factor is (), their least common multiple is (). Explain the reason

In Division A, if the divisor is divided by the divisor, the quotient obtained is all natural number without remainder, that is to say, the divisor is the multiple of the divisor, and the divisor is the factor of the divisor. B we divide a composite number into several prime numbers, which are called the prime factors of the composite number
The greatest common factor is 1, the greatest common factor of the number of Coprime relation is only 1, the least common multiple is ab, the least common multiple of the number of Coprime relation is their product
A and B are coprime numbers, their greatest common factor is (1), and their least common multiple is (AB). Because a and B are coprime numbers with only one factor and itself, the common multiple is their product.
The textbook defines coprime numbers as follows: some natural numbers whose greatest common factor is 1 are called coprime numbers. Two numbers are the greatest common divisor, and only two numbers of 1 are coprime numbers. So the greatest common factor is 1. Since there are no other common factors, the least common multiple is their product a * B
A and B are coprime numbers, their greatest common factor is (1), and their least common multiple is (AB). Because a and B are coprime numbers with only one factor and itself, the common multiple is their product.
A and B are coprime numbers, and their greatest common factor is______ The least common multiple is______ .
Because a and B are coprime numbers, their greatest common factor is 1 and their least common multiple is ab
If the complex z = (I1 − I) 2, then the corresponding point of the complex Z + 1 on the complex plane is ()
A. First quadrant B. second quadrant C. third quadrant D. fourth quadrant
∵ z = (I1 − I) 2 = [I (1 + I) (1 − I) (1 + I)] 2 = (− 1 + I2) 2 = (− 1 + I) 24 = − 2i4 = − 12I, ∵ Z + 1 = 1-12i. ∵ the coordinates of the point corresponding to the complex Z + 1 in the complex plane are (1, − 12), which is located in the fourth quadrant
High middle polar coordinates and parametric equation formula
x = r*cos(θ),
y = r*sin(θ),
x=Psina
y=Pcosa
P is the polar axis
A is the polar angle
x=r*sin(theta)
y=r*cos(theta)
r=sqrt(x^2+y^2)
theta=atan(y/x)
x=rsin a,
y=rcos a,
Zero
Let the complex number Z = (m-1) + (M & # 178; - 4m-5) I correspond to the point Z in the complex plane. If the position of the point Z satisfies the following requirements respectively, find the conditions that the real number m satisfies:
(1) Not on the real axis (2) on the imaginary axis (3) below the real axis (excluding the real axis) (4) to the right of the imaginary axis (excluding the imaginary axis)
(1) Not on the real axis,
(M & # 178; - 4m-5) ≠ 0, the solution is m ≠ 5, m ≠ - 1
(2) On the imaginary axis
(m-1) = 0, the solution is m = 1
(3) Below the real axis (excluding the real axis)
(m²-4m-5)
For example, if there are any parts that do not change or change, they are all parts of the filter, and all the parts need to be cleaned
Mathematical problems about polar coordinates and parametric equations in senior one
Known circle (x-2cosa) ^ 2 + (y + 2cos2a-2) ^ 2 = 1
(1) Find the trajectory equation of the center of a circle
(2) If there is a linear equation with P (0, b) intersecting at points a and B, and | PA |, | ab |, | Pb |, the value range of B is obtained
(1) (2) if | PA |, | ab |, | Pb | form an equal ratio sequence, then | ab |, | Pb |; = | PA | * | Pb |
Let the set P = {- 4, - 3, - 2,0}, q = {0,1,2}, take a random number from the set P as X, and a random number from the set Q as Z, and calculate the probability that the complex Z is a pure imaginary number
Take n = 4 * 3 = 12 methods
Z is a pure imaginary number, so x = 0, y is not 0
There are 1 * 2 kinds
So the probability is 2 / 12 = 1 / 6
High school mathematics problems (polar equation)
It is known that the rectangular coordinate equation of parabola is y ^ 2 = 4x
(1) Establish the appropriate polar coordinate system, write the parabolic polar coordinate equation
The answer to this question is to focus on the pole, and finally get ρ = 2 / (1-cos θ). What's the matter?
Taking the origin as the pole, the positive half axis of X is the polar day
Then y = ρ sin θ, x = ρ cos θ
So ρ ^ 2Sin ^ 2 θ = ρ cos θ
ρsin^2 θ=cosθ
Let the set P = {- 4, - 3, - 2,0}, q = {0,1,2}, take a random number from the set P as X, and a random number from the set Q as Z, and calculate the probability that the complex Z is a pure imaginary number
Take a random number from the set Q as y?
Z is a pure imaginary number, then x = 0 and Y ≠ 0
The probability of X being 0 is 1 / 4, and the probability of y not being 0 is 2 / 3
Then the probability that z is an imaginary number is: 1 / 4 × 2 / 3 = 1 / 6
Polar coordinates and equations
Let the polar coordinate equation of curve C be p = 2, the parameter equation of line l be x = t, y = T-2, radical 2, (t is the parameter), then the rectangular coordinate of the intersection of curve C and line L is?
In particular, I can't understand how to translate P = 2 Thank you
If x = ρ cos θ, y = ρ sin θ, (0 ≤ θ < 2 π), then cos θ = x / ρ, sin θ = Y / ρ, and the sum of the squares of the two formulas is: 1 = (COS θ) ^ 2 + (sin θ) ^ 2 = (x / ρ) ^ 2 + (Y / ρ) ^ 2 = (x ^ 2 + y ^ 2) / ρ ^ 2
P represents the polar diameter in polar coordinates, which is equivalent to the radius of the circle in rectangular coordinates P = 2, which is equivalent to x ^ 2 + y ^ 2 = 2 under P = root sign. In the linear equation, replace T with y = X-2 times root sign 2. Synthesize two formulas to solve the system of equations x ^ 2 + (X-2) ^ 2 = 4, and get the number by yourself
p=xcosa
p=ysina
so,,.P^2=x^2+y^2
P = 2 is p ^ 2 = 4
x^2+y^2=4 ①
So.. curve C is a circle with center (0,0) and radius 2
X=t
Y = T-2 radical 2
50: Y = X-2 radical 2
LIANLI
The point of intersection
"Especially I can't understand how to make p = 2"
You have to know that P = xcosa... Unfolds
p=xcosa
p=ysina
so,,.P^2=x^2+y^2
P = 2 is p ^ 2 = 4
x^2+y^2=4 ①
So.. curve C is a circle with center (0,0) and radius 2
X=t
Y = T-2 radical 2
50: Y = X-2 radical 2
LIANLI
The point of intersection
"Especially I can't understand how to make p = 2"
You have to know that P = xcosa P = ysina is easy to do after flexible changes. Do more questions like this. Put it away
P = 2 is a circle with radius 2 in polar coordinate system, and the equation in rectangular coordinate system is x ^ 2 + y ^ 2 = 4
The following is to find the intersection of a circle and a straight line, that is, to solve the following equations:
x^2+y^2=4
Y = X-2 radical 2
Let's figure it out for ourselves