Question

Tính B =.\(\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{11+2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)}}\)

Answer 1

\(B=\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{6+3+2+2\sqrt{6}+2\sqrt{12}+2\sqrt{18}}}=\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)^2}}=\frac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)

\(=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+\sqrt{6}+3\sqrt{2}+3}{\sqrt{6}+\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+\sqrt{3}\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)

\(=\frac{\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+1\right)}{\sqrt{6}+\sqrt{3}+\sqrt{2}}=\sqrt{3}+1\)