If a and B are coprime numbers, then their greatest common factor is () and their least common multiple is ()

If a and B are coprime numbers, then their greatest common factor is () and their least common multiple is ()

If a and B are coprime numbers, then their greatest common factor is (1), and their least common multiple is (AB)
If a = 13, B = 14, the least common factor is 1, and the least common multiple is ab
The greatest common factor is (1), and the least common multiple is (AB)
If a and B are coprime numbers, then their greatest common factor is (1), and their least common multiple is (AB)
The greatest common factor is 1 and the least common multiple is a * B.
There is a row of street lamps. The original distance between two adjacent lamps is 8 meters, but now the distance between two adjacent lamps is 12 meters. If one lamp does not move, at least how many meters apart can one lamp not move?
Because the least common multiple of 8 and 12 is 24, at least another one doesn't need to move every 24 meters. A: at least another one doesn't need to move every 24 meters
The greatest common factor and the least common multiple of grade five~
1. A box of apples, nine in nine, seven more; 12 in 12, ten more. How many apples is there in this box at least?
2. There are 36 pieces of chocolate and 39 pieces of sugar, which are distributed to several children on average. As a result, there is one more piece of chocolate and one less piece of sugar. How many children are these chocolates and sugar distributed to at most?
3. From one end of the venue to the other end, the total length is 100 meters. From the first section, a red flag is inserted every 4 meters. Now it is necessary to insert a red flag every 5 meters. How many red flags do not have to be pulled out?
4. On each side of the 60 meter long and 54 meter wide rectangular flower bed, the peach trees are required to be planted at four vertices, and the spacing between any two adjacent trees should be the same, at least how many trees should be planted?
As soon as possible, in the one hour class. The answer must be careful, but also have the formula!
1.34 is 9, the number of 9 is less than 2, the number of 12 is less than 2. The least common multiple of 9 and 12 is 36, plus 2, that is 34.2.5, that is, 35 pieces of chocolate is just enough, 40 pieces of sugar is just enough. The greatest common divisor of 35 and 40 is 5.3.6, and the multiple of 20 need not be pulled. There are 0, 20, 40, 60, 80100.4.38 trees
In the complex plane, the point corresponding to the complex Z satisfying the condition Z · (1 + I) = 2 is located in ()
A. First quadrant B. second quadrant C. third quadrant D. fourth quadrant
∵ Z · (1 + I) = 2, ∵ Z · (1 + I) (1-I) = 2 (1-I). ∵ z = 1-I, the point corresponding to complex Z is (1, - 1)
It is known that the equation of x 2 - (6 + I) x + 9 + AI = 0 (a ∈ R) has real root B. (1) find the value of real numbers a and B. (2) if the complex Z satisfies |. Z-a-bi | - 2 | Z | = 0, then | Z | has the minimum value and the value of | Z |
(1) ∵ B is the real root of the equation X2 - (6 + I) x + 9 + AI = 0 (a ∈ R); (b2-6b + 9) + (a-b) I = 0, ∵ B2 − 6B + 9 = 0A = B, the solution of which is a = b = 3. (2) let z = x + Yi (x, y ∈ R), from |. Z-3-3i | = 2 | Z |, we get (x-3) 2 + (y + 3) 2 = 4 (x2 + Y2), that is, (x + 1) 2 + (Y-1) 2 =
How to read the s after the plural in English?
In general, add - S 1. Clear consonants after reading / S /; map maps 2. Voiced consonants and vowels after bag bags reading / Z /; car cars words ending with s, SH, CH, X and so on add - es reading / iz /
Mathematical complex number of senior two
On the equation x ^ 2 - (tanw + I) x - (2 + I) = 0 w ∈ r of X
It is proved that there is no pure imaginary solution for any real number W primitive equation
2 if the equation has an imaginary root 2 + I, find the other root and the acute angle of W
(1) Let x = Bi (B ≠ 0) (BI) &# 178; - (tanw + I) (BI) - (2 + I) = 0-B & # 178; - (btanw) i-bi & # 178; - 2-I = 0 (- B & # 178; + b-2) - (btanw + 1) I = 0-B & # 178; + B-2 = 0 and btanw + 1 = 0, where - B & # 178; + B-2 = 0 is B & # 178; - B + 2 = 0 because Δ = 1-8
Given that a ＞ 0, B ＞ 0, the equation x × (a ＞ BI) x × (a ＞ 1 ＞ AI = 0 has real roots, find the minimum value of a, and solve the equation when a takes the minimum value
We get the following equation: (a + 2) + (x),
Given that the equation has real roots, we can get x ^ 2 + ax + 1 = 0 and BX + a = 0
From BX + a = 0, we can get: x = - A / B,
Substituting x ^ 2 + ax + 1 = 0, we get B ^ 2-A ^ 2B + A ^ 2 = 0;
If the equation considered as B has real roots, then the discriminant = a ^ 4-4a ^ 2 ≥ 0,
When a > 0 is known, the solution is a ≥ 2;
The minimum value of a is 2
B ^ 2-4b + 4 = 0, B = 2;
Therefore, the solution of the equation is x = - A / b = - 1
Two
English words for teeth
Tooth tooth complex
Given the complex z = (a ^ 2-a-2) + (a ^ 2-a-6) I (a belongs to R), then when a is a value, the point corresponding to Z is below the line x + y = 0
Below the line x + y = 0 is x + y < 0
That is to say, to solve the inequality a ^ 2 - A - 2 + A ^ 2-a-6 < 0
Let the real part be the opposite of the imaginary part