Answer
See the explanation
Work Step by Step
$\begin{array}{llll}x & 0& 1&2& 3& 4& 5&6\\
\frac{x-3}{x^2-x-6} & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & Undefined& \frac{1}{6} & \frac{1}{7}& \frac{1}{8}\\
\frac{1}{x+2} & \frac{1}{2} & \frac{1}{3}& \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8}
\end{array}$
- In conclusion, for $x\ne3$, the rational expression $\frac{x-3}{x^2-x-6} $brought the same result as that of $\frac{1}{x+2}$.
