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(B) Let $y = \frac{x^2-5x+7}{x^2-5x-7}$.Let $t = x^2-5x$. Then $y = \frac{t+7}{t-7}$.Since $x^2-5x = (x-2.5)^2 - 6.25$,the minimum value of $t$ is $-6

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$1$

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(B) Let $y = \frac{x^2-5x+7}{x^2-5x-7}$.Let $t = x^2-5x$. Then $y = \frac{t+7}{t-7}$.Since $x^2-5x = (x-2.5)^2 - 6.25$,the minimum value of $t$ is $-6.25$. Thus $t \in [-6.25, \infty)$.We have $y = \frac{t-7+14}{t-7} = 1 + \frac{14}{t-7}$.For $t \in [-6.25, \infty)$,$t-7 \in [-13.25, \infty)$.As $t 	o 7$,$y 	o \pm \infty$. The function covers all values except the horizontal asymptote $y=1$.For $t \in [-6.25, 7)$,$t-7 \in [-13.25, 0)$,so $y \in (-\infty, 1 + \frac{14}{-13.25}] = (-\infty, -0.056]$.The greatest negative integer in this range is $-1$.For $t \in (7, \infty)$,$t-7 \in (0, \infty)$,so $y \in (1 + 0, \infty) = (1, \infty)$.The least positive integer in this range is $2$.The sum is $-1 + 2 = 1$.

The sum of the least positive integer and the greatest negative integer in the range of the function $f(x) = \frac{x^2-5x+7}{x^2-5x-7}$ is

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