b,
$(2.\sqrt[]8-3.\sqrt[]3+1):\sqrt[]6$
$=\dfrac{2.\sqrt[]8}{\sqrt[]6}-\dfrac{3.\sqrt[]3}{\sqrt[]6}+\dfrac{1}{\sqrt[]6}$
$=\dfrac{\sqrt[]{2.8}}{\sqrt[]3}-\dfrac{3}{\sqrt[]2}+\dfrac{1}{\sqrt[]6}$
$=\dfrac{4}{\sqrt[]3}-\dfrac{3}{\sqrt[]2}+\dfrac{1}{\sqrt[]6}$
c,
$(\dfrac{1}{2}.\sqrt[]{\dfrac{1}{2}}-\dfrac{3}{2}.\sqrt[]{4,5}+\dfrac{2}{5}.\sqrt[]{50}):\dfrac{4}{15}.\sqrt[]{\dfrac{1}{8}}$
$=(\sqrt[]{\dfrac{1}{8}}-\dfrac{3}{2}.\sqrt[]{\dfrac{9}{2}}+\dfrac{2}{5}.\sqrt[]2.5):\dfrac{4}{15}.\sqrt[]{\dfrac{1}{8}}$
$=(\sqrt[]{\dfrac{1}{8}}-9.\sqrt[]{\dfrac{1}{8}}+8.\sqrt[]{\dfrac{1}{8}}:\dfrac{4}{15}.\sqrt[]{\dfrac{1}{8}}$
$=\sqrt[]{\dfrac{1}{8}}(1-9+8):\dfrac{4}{15}.\sqrt[]{\dfrac{1}{8}}$
$=0:\dfrac{4}{15}$
$=0$
