Answer
$\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{x-4\sqrt{x}+4}{x-4}$
Work Step by Step
$\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify:
$\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-2}=...$
$...=\dfrac{(\sqrt{x}-2)^{2}}{(\sqrt{x})^{2}-2^{2}}=\dfrac{(\sqrt{x})^{2}-4\sqrt{x}+2^{2}}{x-4}=...$
$...=\dfrac{x-4\sqrt{x}+4}{x-4}$
