## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}=3\sqrt{2}-2\sqrt{3}$
$\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$ (rationalize the denominator) Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible: $\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=...$ $...=\dfrac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\dfrac{\sqrt{18}-\sqrt{12}}{3-2}=\sqrt{18}-\sqrt{12}=...$ $...=3\sqrt{2}-2\sqrt{3}$