Question

a)\(\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)^2-\left(\sqrt{3}+\sqrt{2}\right)\)

b)\(\dfrac{1}{7+4\sqrt{4}}+\dfrac{1}{7-4\sqrt{3}}\)

Answer 1

\(\text{a) }\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)^2-\left(\sqrt{3}+\sqrt{2}\right)\\ =\sqrt{3}\left(2-2\sqrt{6}+3\right)-\left(\sqrt{3}+\sqrt{2}\right)\\ =5\sqrt{3}-2\sqrt{18}-\sqrt{3}-\sqrt{2}\\ =4\sqrt{3}-6\sqrt{2}-\sqrt{2}\\ =4\sqrt{3}-7\sqrt{2}\)

\(b\text{) }\dfrac{1}{7+4\sqrt{4}}+\dfrac{1}{7-4\sqrt{3}}\\ =\dfrac{1}{7+8}+\dfrac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\\ =\dfrac{1}{15}+\dfrac{7+4\sqrt{3}}{49-48}\\ =\dfrac{1}{15}+7+4\sqrt{3}\)