Question

rút gọn M =\(\sqrt{6+2\sqrt{4-\sqrt{12}}}\)

           N =\(\sqrt{6-2\sqrt{4+\sqrt{12}}}\)

Chứng minh M.N= M+N

Answer 1

M + N = \(\sqrt{6+2\sqrt{4-\sqrt{12}}}+\sqrt{6-2\sqrt{4+\sqrt{12}}}\)

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

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

             = \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\left(1\right)\)

MN = \(\sqrt{6+2\sqrt{4-\sqrt{12}}}\sqrt{6-2\sqrt{4+\sqrt{12}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)

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

      \(=\sqrt{16-12}=\sqrt{4}=2\left(2\right)\)

Từ (1) và (2) => ???