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

Answer 1

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

Answer 2

Wrong