The greatest common divisor and the least common multiple of 12, 24 and 46

The greatest common divisor and the least common multiple of 12, 24 and 46

12=2×6
24=2×12
46=2×23
Obviously, the greatest common divisor of 12, 24 and 46 is 2
∵24=12×2
The least common multiple of 12, 24 and 46 is the least common multiple of 24 and 46
From 24 = 2 × 12; 46 = 2 × 23
The least common multiple of 24 and 46 is 2 × 12 × 23 = 552
The least common multiple of 12, 24 and 46 is 552

The greatest common factor and the least common multiple of 23 and 15

The greatest common factor is 1 and the least common multiple is 23 * 15 = 345

What are the greatest common factor and the least common multiple of 23 and 6

Least common multiple: 138
Greatest common factor: 1

Second grade mathematics fractional equation problem solving!
When m is the value, the equation 2 / x + 1 + 5 / 1-x = m / X has no solution
Solving Tat

Your topic should be 2 / (x + 1) + 5 / (1-x) = m / X!

If one of the two equations x2 + (K-2) x + 2k-1 = 0 is between 0 and 1, and the other is between 1 and 2, then the value range of real number k is______ ．

Let f (x) = x2 + (K-2) x + 2k-1 ∵ of the two equations x2 + (K-2) x + 2k-1 = 0, one is between 0 and 1, the other is between 1 and 2, ∵ f (0) > 0, f (1) < 0, f (2) > 0 ∵ 2K − 1 > 03k − 2 < 04k − 1 > 0 ∵ 12 < K < 23 ∵ the value range of real number k is (12, 23), so the answer is: (12, 23)

It is known that the solution of the equation 2 (x + k) - 1 = 0 about X is one-third smaller than that of 4x-k = 2K + 1. Find the value of the square-k-1 of the algebraic formula K
Seeking the right

2(x+k)-1=0
x+k=1/2
x=1/2-k
4x-k=2k+1
4x=3k+1
x=(3k+1)/4
According to the meaning of the title, we can draw a conclusion
1/2-k+1/3=(3k+1)/4
5/6-k=3k/4+1/4
7k/4=5/6-1/4
7k=10/3-1
k=1/3

Let someone swim along a river, LM, and then swim upstream to the starting point. Let the swimming speed of the person in still water be XM / s, and the current speed be nm / s, and find the time t required for him to go back and forth

L/(x+n) +L/(x-n)

On the application of fractional equation
A new kind of coating is prepared with 100 yuan a coating and 240 yuan B coating. The price per kilogram of the new coating is 3 yuan less than that of a coating and 1 yuan more than that of B coating. How much is the price per kilogram of all kinds of new coatings?

Suppose the price of the new paint is x yuan per kilogram,
100/(x+3)+240/(x-1)=(100+240)/x
At the same time, multiply (x + 3) (x-1) x to get
100(x-1)x+240（x+3）x=340（x+3）(x-1)
100x2-100x + 240x2 + 720x without brackets = 340x2 + 680x-1020
60x = 1020 for items of the same kind
x=17

Solving the fractional equation: x-4 / 5-x - 3 = negative 1 / X-5

Variant
X-4/5-X -3 =1/5-X
Multiply by 5-x
X-4-3 (5-x) = 1
solve equations
4X=20
X=5

On the fractional equation of X: the solution of XX + 1 = 12 is______ ．

XX + 1 = 12, the two sides of the equation multiply by 2 (x + 1) to get 2x = x + 1, and the solution is x = 1. Test: substitute x = 1 into 2 (x + 1) = 4 ≠ 0