The divisor of 16 is______ Among these divisors, four divisors are selected to form at least two proportions______ ．

The divisor of 16 is______ Among these divisors, four divisors are selected to form at least two proportions______ ．

The divisors of 16 are: 1, 2, 4, 8, 16; 1:2 = 8:16, 2:1 = 16:8 (the answer is not unique);

The factor of 18 is______ From which four numbers are selected to form a proportion______ ．

The factors of 18 are: 1, 2, 3, 6, 9, 18; 1:2 = 3:6; or 3:9 = 6:18; (the answer is not unique). So the answer is: 1, 2, 3, 6, 9, 18; 1:2 = 3:6

Write a ratio of 18______ ．

(1) The divisors of 18 are: 1, 2, 3, 6, 9, 18; (2) 2 × 9 = 1 × 18, 2 and 9 as internal terms, 1 and 18 as external terms, the proportion is 1:2 = 9:18; so the answer is: 1:2 = 9:18

Choose four numbers from the divisors of 30 to form a proportion______ ．

The divisors of 30 are: 1, 30, 2, 15, 3, 10, 5, 6; then 3:30 = 1:10

I is an imaginary unit, Z1 = 2 + 3I, Z2 = - 1 + 5I, if Z3 × Z1 = Z2
(1) Find the complex number Z3; (2) if Z3 is a complex root of the equation AX ^ 2 + BX + 2 = 0, find another root of the equation and the values of real numbers a and B

1. Z3 = Z2 / Z1 = (- 1 + 5I) / (2 + 3I) = (- 1 + 5I) (2-3i) / (2 + 3I) (2-3i) = (- 2 + 3I + 10I + 15) / (4 + 9) = 1 + I2, AB is a real number, then the two imaginary roots are conjugate imaginary numbers, so the other root is 1-i

Let Z1 = 1 + 2ai, Z2 = A-I (a ∈ R), given a = {Z | z-z1 | ≤ 1}, B = {Z | z-z2 | ≤ 2}, a ∩ B = empty set, find the value range of A

Let z = x + Yi, then: z-z1 = (x-1) + (y-2a) I, z-z2 = (x-a) + (y + 1) I, then: | z-z1 | = √ [(x-1) & sup2; + (y-2a) & sup2;] "1, | z-z2 | = √ [(x-a) & sup2; + (y + 1) & sup2;]" 2 "in analytic geometry, we know: (x-1) & sup2; + (y-2a) & sup2;" 1 table

It is known that complex numbers Z1 and Z2 satisfy Z1 ^ 2 = Z2, if Z1 = 1 + mi (M belongs to R), | Z1|

This paper will be the first and last (1 + m \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\inthe case of + 12) ∵ M & # 178; ≤ 1  √ 13 ≤ | Z2 + 3 | ≤ 4 | Z2 + 3 |

The least common multiple of three primes is 70. What is the sum of the three primes
Write the name of the note

When we decompose 70 into prime factors, we get all prime numbers, and there must be three, because one more will lead to composite numbers. When we decompose 70 into prime factors, 70 = 2 * 5 * 7
So these three numbers are 2, 5 and 7 respectively
The sum is 2 + 5 + 7 = 14

The least common multiple of three prime numbers is 399. What are the three numbers?

Directly decompose 399 into prime factor
The answer is 3, 7 and 19

The least common multiple of three different prime numbers is 154. Find these three numbers

2、7、11