Answer
$\dfrac{-7}{\sqrt{x}-3}=-\dfrac{7(\sqrt{x}+3)}{x-9}$
Work Step by Step
$\dfrac{-7}{\sqrt{x}-3}$
Multiply the numerator and the denominator by the conjugate of the denominator and simplify:
$\dfrac{-7}{\sqrt{x}-3}=\dfrac{-7}{-3+\sqrt{x}}\cdot\dfrac{-3-\sqrt{x}}{-3-\sqrt{x}}=\dfrac{-7(-3-\sqrt{x})}{(-3)^{2}-(\sqrt{x})^{2}}=...$
$...=\dfrac{-7(-3-\sqrt{x})}{9-x}=\dfrac{7(\sqrt{x}+3)}{9-x}=-\dfrac{7(\sqrt{x}+3)}{x-9}$
