[tex]\bf 5. \\ \\ \it \ \ \Rightarrow a=\dfrac{1}{\sqrt3}(1-1+\dfrac{1}{3})=\dfrac{1}{3\sqrt3}\\ \\ \\ \Rightarrow b=\dfrac{\sqrt6}{\sqrt8}(1+\dfrac{3}{2})=\dfrac{\sqrt3}{2}\cdot\dfrac{5}{2}=\dfrac{5\sqrt3}{4}\\ \\ \\ a\cdot b=\dfrac{1}{3\sqrt3}\cdot\dfrac{5\sqrt3}{4}=\dfrac{5}{12}[/tex]
7.
[tex]\it x\sqrt{27}-\sqrt{12}=\sqrt{300}|_{:\sqrt3}\ \Rightarrow 3x-2=10|_{+2} \Rightarrow 3x=12|_{:3} \Rightarrow x=4[/tex]
