A = a * 2 * 3, B = a * 3 * 3 (a is prime), then the greatest common factor of a and B is (), and the least common multiple is ()

A = a * 2 * 3, B = a * 3 * 3 (a is prime), then the greatest common factor of a and B is (), and the least common multiple is ()

The greatest common factor is 3a, and the least common multiple is 18a. Just give an example!
3a,18
The greatest common factor of a and B is (3a), and the least common multiple is (18a)
A = a × B, B = C × A. (a, B, C are different prime numbers), the greatest common divisor of a and B is (a), and the least common multiple is ()
A = a × B, B = C × A. (a, B, C are different prime numbers), the greatest common divisor of a and B is (a), and the least common multiple is (a × B × C)
ABC
Does family have a plural? If so, what is the plural?
When family means family, we can't use the plural; but if family means family, we can use the plural
The plural is: families
It is known that z = I + 1 is a root of the equation Z ^ 2 + AZ + B = 0
1. Find the value of real numbers a and B
2. Conjecture another root of the equation and prove it by combining WIDA's theorem
(1) Substituting: I ^ 2 + 2I + 1 + AI + A + B = 0, so (a + b) + (a + 2) I = 0
So a = - 2, B = 2
(2) Another root = - A-Z = 1-I
Family plural
families
Families, this word is very simple
If the L before y is not a vowel, change y to I plus es
If y is preceded by a vowel, add es directly
families
families
Because the L before y is not a vowel
So we need to change y to I plus es
families
families
Go to y plus y
=families
famiies
families
Because the L before y is not a vowel
So we need to change y to I plus es
If y is preceded by a vowel, add s directly
For example: Monkeys
families
families
Change y to I + es
families。。。。。。。。。。。。。。。。。。。。
U weak or you delivery time will certainly pay time delivery speed recovery
families
familys
families
We know the equations about x.y
（2x-1）+i=y-（3-y）i
A kind of
(2x + ay) - (4x-y + b) I = 9-8i has real number solution
Find the values of real numbers a and B
Comparing the two sides, we can see that the real part equals the real part and the imaginary part equals the imaginary part
2x-1=y
3-y=-1
We get x = 5 / 2, y = 4
Substitute the following equation
2x+ay=9
4x-y+b=8
We get a = 1, B = 2
1) Equation 2x-3 + I = y - (3-y) I = > y = 4, x = 2.5
Substituting into formula 2)
2*（2.5）+4a=9=>a=1
4*(2.5)-4+b=8=>b=2
∵x,y,a,b∈R
（2x-1）+i=y-（3-y）i
The real part and the imaginary part are equal respectively
Then 2x-1 = y
1=-(3-y)
Solution
y=4,x=5/2
In the same way
(2x-ay)-(4x-y+b)i=9-8i
That is, (5-4a) - (6 + b) I = 9-8i
5-4a=9
-(6-b)=-8
So a = - 1, B = - 2
If the complex number m + 2i1 − I (m ∈ R, I is an imaginary unit) is a pure imaginary number, then M=______ .
M + 2i1 − I (M + 2I) (1 + I) (1 − I) (1 + I) = m − 22 + m + 12I. According to the concept of pure imaginary number, m − 22 = 0 & nbsp; m + 12 ≠ 0, the solution is m = 2
A question about high school plural, solving, thank you,
2. Known equations about x.y
（ix-1）+i=y-（3-y）i
A kind of
(IX ay) = (4x-y + b) I = 9-8i has real solution
Find the values of real numbers a and B
Want a detailed problem-solving process
Thank you for your help,
The equations are written incorrectly and correctly as follows,
（2x-1）+i=y-（3-y）i
A kind of
（2x+ay）-（4x-y+b）i=9-8i
There are real solutions, that is, X and y are real numbers
So x, y, a, B are all real numbers
（2x-1）+i=y-（3-y）i
The real part and the imaginary part are equal respectively
2x-1=y
1=-(3-y)
y=4,x=5/2
(2x-ay)-(4x-y+b)i=9-8i
That is, (5-4a) - (6 + b) I = 9-8i
5-4a=9
-(6-b)=-8
So a = - 1, B = - 2
According to the first equation:
2x-1+i=y-（3-y）i
Because the equation has real solutions
2x-1=y；
1=-（3-y）
So y = 4; X = 5 / 2
Bring it into the second equation, and use the conditions of a and B real numbers, the method is the same as above
The results are as follows
2x-ay=9
4x-y+b=8
So it can be concluded that
a=-1
B=2
1. One cube root of complex number - I is I, and the other two cube roots are?
2. Square root of ball 7-24i
-1 / 2 + I * (radical 3 / 2)
-1 / 2-I (radical 3 / 2)
Let the square root of 7-24i be x + iy
Then (x + iy) * (x + iy) = 7-24i
have to
X2-Y2=7
2XY=-24
Solve XY
1. When I get home from work, I'll give you a reply today.
1 (root 3-I) / 2 (- root 3-I) / 2
2 4-3i -4+3i
Let the cube root of - I be a, R be the module of a, and X be the argument of A
A ^ 3 = [R (cosx + isinx)] ^ 3 = - I = cos (tt / 2) - I * sin (tt / 2), i.e
r^3(cos3x+isin3x)=cos(3TT/2)+isin(3TT/2)
So, r = 1,3x = 3tt / 2 + 2ktt
When k = 0, x = tt / 2, ---- a = costt / 2 + isintt / 2 = I
K =... Expand
Let the cube root of - I be a, R be the module of a, and X be the argument of A
A ^ 3 = [R (cosx + isinx)] ^ 3 = - I = cos (tt / 2) - I * sin (tt / 2), i.e
r^3(cos3x+isin3x)=cos(3TT/2)+isin(3TT/2)
So, r = 1,3x = 3tt / 2 + 2ktt
When k = 0, x = tt / 2, ---- a = costt / 2 + isintt / 2 = I
When k = 1, x = tt / 2 + 2tt / 3, - a = cos (tt / 2 + 2tt / 3) + isin (tt / 2 + 2tt / 3) =?
When k = 2, x = tt / 2 + 4tt / 3, - a = cos (tt / 2 + 4tt / 3) + isin (tt / 2 + 4tt / 3) =?
2. Two methods
i. Using the algebraic form of complex number:
Let the square root be a + bi, then
(a + bi) ^ 2 = 7-24i, i.e
If a ^ 2-B ^ 2 + 2abi = 7-24i, the virtual and real parts are equal
A ^ 2-B ^ 2 = 7,2ab = - 24, simultaneous can get a, B, solve two groups on the line, the redundant rounding. Note that a, B is not virtual
II. Use the plural exponential form:
Let the square root be r (cosx + isinx), then
R ^ 2 (cos2x + isin2x) = 7-24i, i.e
R ^ 2 = under root (7 ^ 2 + 24 ^ 2) ---- modules on both sides are equal
2X = arc?? + 2ktt, is the number on the right, the argument of 7-24i. It is expressed by the inverse trigonometric function
K = 0,1 corresponds to two spokes, so it corresponds to two square roots
High school plural exercises
What is the conjugate complex number of 1 / (1-I) and how to get it
In addition, if w = - 1 / 2 + radical 3 / 2I, then 1 + W =?
Please pay for the process, thank you
1/(1-i)
=(1+i)/[(1-i)(1+i)]
=(1-i)=(1+1)
=1/2-i/2
So the conjugate complex number is 1 / 2 + I / 2
1+w=1-1/2+√3/2i=1/2+√3/2i
For example, the conjugate of 1 + 2I is 1-2i
The second question is also rational denominator, you know (a-b) (a + b) = A2-B2