The least common multiple of two different primes must be a composite number______ (judge right or wrong)

The least common multiple of two different primes must be a composite number______ (judge right or wrong)

The analysis shows that the least common multiple of two different prime numbers must be a composite number
A = 2 × 3 × N2, B = 3 × N3 × 5, (n is prime), then the greatest common divisor of a and B is______ The least common multiple is______ .
A = 2 × 3 × N2, B = 3 × N3 × 5 (n is prime), so the greatest common divisor of a and B is 3 × N2; the least common multiple of a and B is 2 × 3 × N3 × 5; so the answer is: 3 × N2, 2 × 3 × N3 × 5
3×n²
2X3×n³×5
The difference between impure imaginary number and complex number?
Complex numbers include real numbers and imaginary numbers, such as a, a + bi, Bi, where a and B are real numbers
Impure imaginary numbers should be in the form of a + bi
------------------------------------------------------------------------------------
It can also be seen from the complex plane that the numbers corresponding to all points on the complex plane are complex numbers (including real numbers and imaginary numbers);
(1) The numbers corresponding to the points on the y-axis (except the origin) are pure imaginary numbers;
(2) The numbers corresponding to the points on the x-axis are all real numbers;
(3) Except (1) (2), the numbers corresponding to the points are all non pure imaginary numbers
The complex number is a further extension of the real number. Including: real number and imaginary number
Imaginary numbers include pure imaginary numbers and non pure imaginary numbers
The scope is not the same, including the classification is not the same
When AB is not zero, I don't think it makes any difference.
Definition of complex number: a + bi
Definition of impure imaginary number: a + bi (where a and B are not zero)
From this point of view, the complex number a + bi is not restricted by a and B. It has three possibilities
1. When a = 0, Bi is a pure imaginary number
2. When B = 0, a is a real number
3. When a ≠ 0, B ≠ 0, it is a non pure imaginary number
In other words, the non pure imaginary number must be a complex number, but the complex number is not necessarily a non pure imaginary number, because the complex number can also be a real number
Definition of complex number: a + bi
Definition of impure imaginary number: a + bi (where a and B are not zero)
From this point of view, the complex number a + bi is not restricted by a and B. It has three possibilities
1. When a = 0, Bi is a pure imaginary number
2. When B = 0, a is a real number
3. When a ≠ 0, B ≠ 0, it is a non pure imaginary number
In other words, a non pure imaginary number must be a complex number, but the complex number is not necessarily a non pure imaginary number, because the complex number can also be a real number
Given the complex number Z = a + bi, if the conjugate complex number of Z + Z and the conjugate complex number of Z * Z are the two roots of the equation x square-3x + 2 = 0, find a, B
The conjugate complex numbers of Z and Z are two of x ^ 2-x + 2 = 0 or x ^ 2-2x + 1 = 0
X = 1 / 2 ± (√ 7 / 2) * I or X1 = x2 = 1
A = 1 / 2, B = ± √ 7 / 2 or a = 1, B = 0
Plural of English words!
Plural of voice
Plural of price
Plural of case
Plural of torch
The plural of box
Plural of country
voices
prices
cases
lorches
boxes
countries
On the complex number of imaginary numbers
1. Given that the complex Z satisfies Z + | Z | = 2 + 8i, find the complex Z
2. Find the value of | Z1 | = 5, | Z2 | = 3, | Z1 + Z2 | = 6
1. Let z = a + biz + | Z | = a + bi + radical (a ^ 2 + B ^ 2) = 2 + 8i, so B = 8, a = - 152. Let Z1 = a + biz2 = C + diz1 + Z2 = (a + C) + (B + D) ia ^ 2 + B ^ 2 = 25 C ^ 2 + D ^ 2 = 9 (a + C) ^ 2 + (B + D) ^ 2 = 36, so 2Ac + 2bd = 36-25-9 = 2z1-z2 = (A-C) + (B-D) I | z1-z2 | = (A-C) ^ 2 + (B-D) ^ 2 = 25 + 9 -
It is proved that the imaginary roots of an equation of degree n with real coefficients appear in pairs, that is, if z = a + bi (B ≠ 0) is a root of the equation, then = a-bi is also a root
I have limited points, so the reward is not too high
Let a * denote the conjugate complex number of a, that is, (a + bi) * = a-bi
There are (AB) * = a * × b *, (a + b) * = a * + b *
Let Z be the solution of ∑ akx ^ k = 0
That is, ∑ AKZ ^ k = 0, (∑ AKZ ^ k) * = 0 * = 0
（∑akz^k）*＝∑[（ak）*×（z^k）*]＝∑ak（z*）^k＝0
(note that AK is a real number, AK * = AK.)
It means that Z * is also the solution of ∑ akx ^ k = 0
Plural English words
water
sheep
fish
Air
countable noun
Plural words:
Apple, orange, eraser, elephant, egg, office work, Islan, umbrella
All you can count is...
Generally, those that can distinguish numbers individually are..
The title of complex number: given that | Z | = √ 20, the complex number (1 + 2I) Z is a pure imaginary number, find the complex number Z
Given the complex number Z = at + bi (a, B belong to real number), if there is a real number T, let z = (2 + 4I / T) - ATI hold, find the value of 2a-b
2+4i/t-ati=at+bi;
(4/t-at-b)i=at-2;
=>at=2,4/t-at-b-0;
=>a=2/t ,b=4/t-2;
2a-b=4/t-(4/t-2)
=2
Two