1. We are asked to determine if the functions $h(x)=\frac{x-3}{\sqrt{x+1}+2}$ and $k(x)=\sqrt{x+1}-2$ are equal. 2. Start by simplifying $h(x)$. Notice that $h(x)$ is a fraction with denominator $\sqrt{x+1}+2$. To simplify, multiply numerator and denominator by the conjugate of the denominator, $\sqrt{x+1}-2$: $$h(x) = \frac{x-3}{\sqrt{x+1}+2} \cdot \frac{\sqrt{x+1}-2}{\sqrt{x+1}-2} = \frac{(x-3)(\sqrt{x+1}-2)}{(\sqrt{x+1}+2)(\sqrt{x+1}-2)}$$ 3. Use the difference of squares formula on the denominator: $$(\sqrt{x+1}+2)(\sqrt{x+1}-2) = (\sqrt{x+1})^2 - 2^2 = (x+1) - 4 = x - 3$$ 4. Substitute the denominator back: $$h(x) = \frac{(x-3)(\sqrt{x+1}-2)}{x-3}$$ 5. For all $x$ such that $x \neq 3$, the $x-3$ terms cancel out: $$h(x) = \sqrt{x+1} - 2$$ 6. This matches exactly $k(x)$, so $h(x) = k(x)$ for all $x \neq 3$ where the expressions are defined. 7. Note, however, at $x=3$ the original function $h(x)$ is undefined (division by zero), but $k(3) = \sqrt{4} - 2 = 2 - 2 = 0$ is defined. 8. Conclusion: The functions $h$ and $k$ are equal on their common domain except at $x=3$ where $h$ is undefined but $k$ is defined. Final answer: $h(x)$ and $k(x)$ represent the same function except at $x=3$ where $h$ is undefined.