\(D=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right).\left(1+\sqrt{2}\right)}=\frac{1}{\sqrt{2}+1}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\)
