Answer
The graphs of both functions are the same.
Work Step by Step
Given
\begin{equation}
\begin{aligned}
& \frac{3}{\sqrt{x+3}-\sqrt{x}}=\sqrt{x+3}+\sqrt{x} \\
& {[0,8,1] \text { by }[0,6,1]}
\end{aligned}
\end{equation}
Let
\begin{equation}
\begin{aligned}
f(x)&= \frac{3}{\sqrt{x+3}-\sqrt{x}}\\
g(x)&=\sqrt{x+3}+\sqrt{x}\\
\end{aligned}
\end{equation}
The graphs of both functions are shown in the figure and are clearly the same. We can prove this by rationalizing the denominator.
\begin{equation}
\begin{aligned}
g(x) & =\frac{3}{\sqrt{x+3}-\sqrt{x}}\cdot \frac{\sqrt{x+3}+\sqrt{x}}{\sqrt{x+3}+\sqrt{x}} \\
& = \frac{3\left( \sqrt{x+3}+\sqrt{x}\right)}{x+3-x}\\
&= \sqrt{x+3}+\sqrt{x}
\end{aligned}
\end{equation}
