Question

Cho N=\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\). Chứng minh rằng N là một số nguyên.

Answer 1

N=\(\sqrt{3+2\sqrt{2}}\)+\(\sqrt{6-4\sqrt{2}}\)

=\(\sqrt{1+2\sqrt{2}+2}\)+\(\sqrt{4-2.2\sqrt{2}+2}\)

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

=1+\(\sqrt{2}\)+2-\(\sqrt{2}\)=3