Question

Answer 1

a: \(\frac{2-\sqrt3}{2\sqrt6}=\frac{\sqrt6\left(2-\sqrt3\right)}{12}=\frac{2\sqrt6-3\sqrt2}{12}\)

b: \(\frac{1}{\sqrt2+\sqrt3}=\frac{1\cdot\left(\sqrt3-\sqrt2\right)}{\left(\sqrt3+\sqrt2\right)\left(\sqrt3-\sqrt2\right)}=\frac{\sqrt3-\sqrt2}{3-2}=\sqrt3-\sqrt2\)

c: \(\frac{-3}{2\sqrt2-3\sqrt3}=\frac{3}{3\sqrt3-2\sqrt2}\)

\(=\frac{3\left(3\sqrt3+2\sqrt2\right)}{\left(3\sqrt3-2\sqrt2\right)\left(3\sqrt3+2\sqrt2\right)}=\frac{3\left(3\sqrt3+2\sqrt2\right)}{19}\)

d: \(\frac{1+\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}+x}{x}\)