Asked by alex
the sum of the reciprocals of 3 consecutive positive integers is 47 divided by the product of the integers. what is the smallest of the 3 integers
Answers
Answered by
Steve
1/(x-1) + 1/x + 1/(x+1) = 47/((x-1)x(x+1))
x(x+1) + (x-1)(x+1) + x(x-1) = 47
x^2 + x + x^2 - 1 + x^2 - x
3x^2 - 1 = 47
3x^2 = 48
x^2 = 16
x = 4
so, the numbers are 3,4,5
x(x+1) + (x-1)(x+1) + x(x-1) = 47
x^2 + x + x^2 - 1 + x^2 - x
3x^2 - 1 = 47
3x^2 = 48
x^2 = 16
x = 4
so, the numbers are 3,4,5
Answered by
bobpursley
Wrong
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.