Answer
$\displaystyle \frac{x-2\sqrt{x}-3}{x-9}$
Work Step by Step
We lose the square roots in the denominator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$
$\displaystyle \frac{\sqrt{x}+1}{\sqrt{x}+3}\color{red}{ \cdot\frac{\sqrt{x}-3}{\sqrt{x}-3} }\qquad$ (rationalize)
$=\displaystyle \frac{(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x})^{2}-(3)^{2}}$
... use FOIL for the numerator
$=\displaystyle \frac{(\sqrt{x})^{2}-3\sqrt{x}+\sqrt{x}-3}{x-9}$
$=\displaystyle \frac{x-2\sqrt{x}-3}{x-9}$
