Question

[Thanh Vu]

a.(\(\sqrt{6}+\sqrt{2}\) ).(\(\sqrt{3}-\sqrt{2}\) ).\(\sqrt{\sqrt{3}+2}\)

b.\(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)

Answer 1

[Ho Truc]

a)\(=\left(\sqrt{2}\cdot\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{\sqrt{3}+2}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{2}\cdot\sqrt{\sqrt{3}+2}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{2\cdot\left(\sqrt{3}+2\right)}\)

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

=\(\left(3-2\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)

=\(\left|\sqrt{3}+1\right|\)=\(\sqrt{3}+1\)