Question

tính

B=\(\sqrt{2+\sqrt{3}}\) . \(\sqrt{2+\sqrt{2+\sqrt{3}}}\) . \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\) . \(\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

Answer 1

Ta có:

\(B=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

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

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

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

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

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

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

Answer 2

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

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

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