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

Wrong

To find the smallest of the three consecutive positive integers, we need to solve the given equation:

Let's assume that the three consecutive positive integers are n, n+1, and n+2.

According to the given statement, the sum of the reciprocals of these three integers is equal to 47 divided by the product of the integers:

1/n + 1/(n+1) + 1/(n+2) = 47/(n * (n + 1) * (n + 2))

Now, to solve this equation, we can start by simplifying the right side:

47/(n * (n + 1) * (n + 2)) = 47/((n^2) + 2n) * (n + 2)

Now, let's cross-multiply to get rid of the fractions:

((n^2) + 2n) * (n + 2) = 47

Expanding the equation:

n^3 + 4n^2 + 4n + 2n^2 + 8n = 47

Combining like terms:

n^3 + 6n^2 + 12n = 47

Rearranging the equation:

n^3 + 6n^2 + 12n - 47 = 0

Now, we need to find the value of n that satisfies this equation. Since this is a cubic equation, finding an exact solution can be difficult. We can use numerical methods, such as factoring or approximation methods, to find an approximate value for n.

Without further information about the value of n, it is not possible to find the smallest of the three integers.

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