Question

a) (\(\sqrt{8}+\sqrt{3}\)).\(\sqrt{6}\)
b) \(\sqrt{4,5}\) - \(\dfrac{1}{2}\).\(\sqrt{72}\)+5.\(\sqrt{\dfrac{1}{2}}\)
c) \(\sqrt{32}\) - \(\sqrt{18}\)+\(\dfrac{4}{\sqrt{2}}\)

Answer 1

a: \(\left(\sqrt{8}+\sqrt{3}\right)\cdot\sqrt{6}\)

\(=\sqrt{8\cdot6}+\sqrt{3\cdot6}\)

\(=\sqrt{48}+\sqrt{18}=4\sqrt{3}+3\sqrt{2}\)

b: \(\sqrt{4,5}-\dfrac{1}{2}\cdot\sqrt{72}+5\cdot\sqrt{\dfrac{1}{2}}\)

\(=\sqrt{\dfrac{9}{2}}-\dfrac{1}{2}\cdot6\sqrt{2}+5\cdot\dfrac{\sqrt{2}}{2}\)

\(=\dfrac{3\sqrt{2}}{2}+\dfrac{5\sqrt{2}}{2}-\dfrac{6\sqrt{2}}{2}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}\)

c: \(\sqrt{32}-\sqrt{18}+\dfrac{4}{\sqrt{2}}\)

\(=4\sqrt{2}-3\sqrt{2}+2\sqrt{2}\)

\(=6\sqrt{2}-3\sqrt{2}=3\sqrt{2}\)