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If $f(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}-1}$, then what is $\frac{f(\mathrm{a})}{f(\mathrm{a}+1)}$ equal to?
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The correct answer is:
$f\left(\mathrm{a}^{2}\right)$
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}-1}$
$\mathrm{f}(\mathrm{a})=\frac{\mathrm{a}}{\mathrm{a}-1}$
$\mathrm{f}(\mathrm{a}+1)=\frac{\mathrm{a}+1}{\mathrm{a}+1-1}=\frac{\mathrm{a}+1}{\mathrm{a}}$
$\therefore \quad \frac{\mathrm{f}(\mathrm{a})}{\mathrm{f}(\mathrm{a}+1)}=\frac{\frac{\mathrm{a}}{\mathrm{a}-1}}{\frac{\mathrm{a}+1}{\mathrm{a}}}=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-1}$
$\mathrm{f}\left(\mathrm{a}^{2}\right)=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-1}$
$\mathrm{f}(\mathrm{a})=\frac{\mathrm{a}}{\mathrm{a}-1}$
$\mathrm{f}(\mathrm{a}+1)=\frac{\mathrm{a}+1}{\mathrm{a}+1-1}=\frac{\mathrm{a}+1}{\mathrm{a}}$
$\therefore \quad \frac{\mathrm{f}(\mathrm{a})}{\mathrm{f}(\mathrm{a}+1)}=\frac{\frac{\mathrm{a}}{\mathrm{a}-1}}{\frac{\mathrm{a}+1}{\mathrm{a}}}=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-1}$
$\mathrm{f}\left(\mathrm{a}^{2}\right)=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-1}$
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