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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

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    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

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