## Precalculus: Mathematics for Calculus, 7th Edition

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