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If ${h(x)={x}^{2}-7x-8}$ and ${j(x)={x}^{2}-1}$, then find $\left(\cfrac{j}{h}\right)(x)$

A

$7x+8$

B

$\cfrac{x-1}{x-8}$

C

$\cfrac{x+1}{x-8}$

D

$\cfrac{x-8}{x-1}$

E

$\cfrac{x-8}{x+1}$

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