Question

Answer 1

4) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}}{2-3}-\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)

\(=-\sqrt{3}-\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(=-2\sqrt{3}\)

5) \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{\sqrt{5}-3}\)

\(=\dfrac{3-\sqrt{5}}{9-5}-\dfrac{\sqrt{5}+3}{5-9}\)

\(=\dfrac{3-\sqrt{5}}{4}+\dfrac{\sqrt{5}+3}{4}\)

\(=\dfrac{3-\sqrt{5}+\sqrt{5}+3}{4}\)

\(=\dfrac{3}{2}=1,5\)

6) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{6}}{2-6}+\dfrac{\sqrt{6}-\sqrt{2}}{6-2}\)

\(=-\dfrac{\sqrt{6}+\sqrt{2}}{4}+\dfrac{\sqrt{6}-\sqrt{2}}{4}\)

\(=\dfrac{-\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{2}}{4}\)

\(=-\dfrac{\sqrt{2}}{2}\)