## Precalculus: Mathematics for Calculus, 7th Edition

$\sqrt{x+1}-\sqrt{x}=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}$
$\sqrt{x+1}-\sqrt{x}$ Multiply this expression by $\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$ and simplify: $\sqrt{x+1}-\sqrt{x}=(\sqrt{x+1}-\sqrt{x})\Big(\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\Big)=...$ $...=\dfrac{(\sqrt{x+1})^{2}-(\sqrt{x})^{2}}{\sqrt{x+1}+\sqrt{x}}=\dfrac{x+1-x}{\sqrt{x+1}+\sqrt{x}}=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}$