The number of positive integers x satisfying | vedclass

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(B) Given equation: $\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x+2} = \frac{13}{12}$.Combining the fractions on the left side:$\frac{(x+1)(x+2) + x(x+2)

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$1$

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(B) Given equation: $\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x+2} = \frac{13}{12}$.Combining the fractions on the left side:$\frac{(x+1)(x+2) + x(x+2) + x(x+1)}{x(x+1)(x+2)} = \frac{13}{12}$.Expanding the numerator:$\frac{(x^2 + 3x + 2) + (x^2 + 2x) + (x^2 + x)}{x(x+1)(x+2)} = \frac{13}{12}$.$\frac{3x^2 + 6x + 2}{x^3 + 3x^2 + 2x} = \frac{13}{12}$.Cross-multiplying:$12(3x^2 + 6x + 2) = 13(x^3 + 3x^2 + 2x)$.$36x^2 + 72x + 24 = 13x^3 + 39x^2 + 26x$.Rearranging into a cubic equation:$13x^3 + 3x^2 - 46x - 24 = 0$.Testing for integer roots using the Rational Root Theorem,we find $x=2$ is a root:$13(8) + 3(4) - 46(2) - 24 = 104 + 12 - 92 - 24 = 0$.Dividing the cubic by $(x-2)$ gives $(x-2)(13x^2 + 29x + 12) = 0$.The quadratic $13x^2 + 29x + 12 = 0$ has discriminant $D = 29^2 - 4(13)(12) = 841 - 624 = 217$,which is not a perfect square,so the roots are not integers.Thus,the only positive integer solution is $x=2$.

The number of positive integers $x$ satisfying the equation $\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x+2} = \frac{13}{12}$ is:

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