Answer
$\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}=2\sqrt{6}-5$
Work Step by Step
$\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible:
$\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\dfrac{(\sqrt{2}-\sqrt{3})^{2}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=...$
$...=\dfrac{(\sqrt{2})^{2}-2(\sqrt{2})(\sqrt{3})+(\sqrt{3})^{2}}{2-3}=\dfrac{2-2\sqrt{6}+3}{-1}=...$
$...=\dfrac{5-2\sqrt{6}}{-1}=2\sqrt{6}-5$
