Answer
$\dfrac{\sqrt{x}+3}{\sqrt{x}}=\dfrac{x-9}{x-3\sqrt{x}}$
Work Step by Step
$\dfrac{\sqrt{x}+3}{\sqrt{x}}$
Multiply the numerator and the denominator of this expression by the conjugate of the numerator and simplify if possible:
$\dfrac{\sqrt{x}+3}{\sqrt{x}}=\dfrac{\sqrt{x}+3}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-3}=\dfrac{(\sqrt{x})^{2}-3^{2}}{\sqrt{x}(\sqrt{x}-3)}=...$
$...=\dfrac{x-9}{\sqrt{x^{2}}-3\sqrt{x}}=\dfrac{x-9}{x-3\sqrt{x}}$
