The least common multiple of two natural numbers is 60. How many possible values are there for the difference between the two natural numbers? twenty-four List all the possibilities and be careful not to repeat them

The least common multiple of two natural numbers is 60. How many possible values are there for the difference between the two natural numbers? twenty-four List all the possibilities and be careful not to repeat them

The least common multiple of two natural numbers is 60. These two natural numbers may be: 60-1 = 59 60-2 = 58 60-3 = 57 60-4 = 56 60-5 = 55 60-6 = 54 60-10 = 50 60-12 = 48 60-15 = 45 60-20 = 40 60-30 = 3060-60 = 0 30-4 = 26 30-12 = 18 30-20 = 10 20-3 = 17 20

It is known that a, B and C are three natural numbers, and the least common multiple of a and B is 60, and the least common multiple of a and C is 270. Find the least common multiple of B and C

When a = 1, B = 60, C = 270, the least common multiple of B and C is 2 × 3 × 5 × 2 × 3 × 3 = 540; when a = 5, B = 2 × 2 × 3 = 12, C = 3 × 3 × 2 = 54, the least common multiple of B and C is 2 × 3 × 2 × 3 × 3 = 108; when a = 2 × 5 = 10, B = 2 × 3 = 6, C = 3 × 3 × 3 = 27, the least common multiple of B and C is 2 × 3 × 3 = 54; when a = 3 × 5, B = 2 × 2 = 4, C = 3 × 3 × 2 = 18, the least common multiple of B and C is 2 × 3 × 3 × 3 = 54 The least common multiple of C is 2 × 3 × 2 × 3 = 36; when a = 2 × 3 × 5, B = 2, C = 3 × 3 = 9, the least common multiple of B and C is 2 × 3 × 3 = 18; answer: the least common multiple of B and C is 540, 108, 54, 36 or 18

(1) The least common multiple of two numbers a and B is 60. How many possible values are there for the difference between the two natural numbers?
If it's short, it's better to write every step clearly. There are also formulas. Don't fill in blindly

There are 23 possibilities
60=2^2*3*5,
A = 60, B can take all the factor formulas of 60: 1,2,3,4,5,6,10,12,15,20,30,60
A = 30, B can take the multiple of all 4 in all factors, a total of 4: 4, 12, 20, 60
A = 20, B can take all the multiples of 3 in all factors, a total of 6: 3, 6, 12, 15, 30, 60
A = 15, B can take the multiple of all 4 in all factors, a total of 4: 4, 12, 20, 60
A = 12, B can take all the multiples of 5 in all factors, a total of 6: 5, 10, 15, 20, 30, 60
A = 10, B can take all the multiples of 12 in all factors, a total of 2: 12,60
A = 6, B can take all multiples of 20 in all factors, a total of 2: 20, 60
A = 5, B can take all the multiples of 12 in all factors, a total of 2: 12,60
A = 4, B can take all the multiples of 15 in all factors, a total of 3: 15, 30, 60
A = 3, B can take all multiples of 20 in all factors, a total of 2: 20, 60
A = 2, B can take all the factors in the multiple of 60, a total of 1: 60
A = 1, B can take all the factors in the multiple of 60, a total of 1: 60
A total of 12 + 4 + 6 + 4 + 6 + 2 + 2 + 2 + 3 + 2 + 1 + 1 = 45 pairs. If we do not consider the order of a and B, there should be 23 cases
(1,60),(2,60),(3,20),(3,60),(4,15),(4,30),(4,60),(5,12),(5,60),(6,20),(6,60),
(10,12),(10,60),(12,15,),(12,20),(12,30),(12,60),
(15,20),(15,60),(20,30),(20,60),(30,60),(60,60)
The difference is 0,2,3,5,7,8,10,11,14,17,18,26,30,40,45,48,50,54,55,56,57,58,59
There are 23 kinds of difference

If the equation x-3:3-2x plus 3-x:2 + MX is equal to - 1 and has an increasing root, then M = what

(3-2x)/(x-30)+(2+mx)/(3-x)=-1
3-2x-2-3-mx=3-x
x(1+m)=-2
There are increasing roots when x = 3
3(1+m)=-2
m=15/3

If the equation X-9 / M + X + 3 / 2 = x-3 / 1 has an increasing root, then the increasing root may be

When denominator X & # 178; - 9 = 0, incrementing root 3 or - 3 may occur
Multiply both sides of the equation by (X & # 178; - 9)
m+2(x-3)=x+3
x=9-m
When m = 6, there is an increasing root x = 3
When m = 12, there is an increasing root x = - 3

If the equation 6 / (x + 1) (x-1) - M / (x-1) = 1 has an increasing root, then what is its increasing root?
A. 0 B, 1 C, - 1 D, 1 and - 1

Multiply by (x + 1) (x-1)
6-m(x+1)=(x+1)(x-1)
If the root is added, the common denominator is 0
(x+1)(x-1)=0
x=-1,x=1
x=-1
Substituting 6-m (x + 1) = (x + 1) (x-1)
6-0=0
Not established
x=1
Substituting 6-m (x + 1) = (x + 1) (x-1)
6-2m=0
m=3
Yes
So the increasing root is x = 1
Choose B

If the fractional equation 2x − 3 = 1 − MX − 3 with respect to X has an increasing root, then the value of M is ()
A. -3B. -2C. -1D. 3

Multiply both sides of the equation by x-3 to get 2 = x-3-m (1). ∵ the original equation has an increasing root, ∵ x-3 = 0, that is, x = 3. Substitute x = 3 into (1) to get m = - 2

How to solve the equation (23 / 3) (5x-16) / 3 = 4x

(23/3)(5x-16)/3=4x
23 (5x-16) = 12x
115x-368=12x
115x-12x=368
103x=368
X = 368 / 103 (about = 3.57)

3 / 5x-19.2 = 2 / 5x how to solve this equation?

3 / 5x-19.2 = 2 / 5x
(3/5-2/5)x=19.2
x/5=19.2
x=19.2*5
x=96

How to solve the equation of 2 / 5x + 3 * 6 = 19?

(2/5)x=19－3×6
(2/5)x=19－18
(2/5)x=1
x=5/2