Question

Answer 1

e) \(=\dfrac{\left(\sqrt{3}-2\right)^2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}=\dfrac{3+4-4\sqrt{3}}{3-4}=-7+4\sqrt{3}\)

f) \(=\dfrac{1+2\sqrt{2}}{1-8}=-\dfrac{1+2\sqrt{2}}{7}\)

g) \(=\dfrac{5\left(\sqrt{6}+\sqrt{3}\right)}{6-3}=\dfrac{5\sqrt{6}+5\sqrt{3}}{3}\)

h) \(=\dfrac{5\left(\sqrt{7}-\sqrt{8}\right)}{7-8}=-5\sqrt{7}+5\sqrt{8}=-5\sqrt{7}+10\sqrt{2}\)

Answer 2

e: \(\dfrac{\sqrt{3}-2}{\sqrt{3}+2}=7-4\sqrt{3}\)

f: \(\dfrac{1}{1-2\sqrt{2}}=\dfrac{-1-2\sqrt{2}}{7}\)