$\sqrt{3}+\sqrt{2}-1$\sqrt{3}+\sqrt{2}-1

$\displaystyle \frac{2+\sqrt{2}-\sqrt{6}}{4}$\displaystyle \frac{2+\sqrt{2}-\sqrt{6}}{4}

$\displaystyle \frac{1}{4}$\displaystyle \frac{1}{4}

$\displaystyle \frac{\sqrt{3}+\sqrt{2}-1}{6}$\displaystyle \frac{\sqrt{3}+\sqrt{2}-1}{6}