Đáp án: $H=1$
Giải thích các bước giải:
Đặt biểu thức trên là $H$
Ta có:
$H=\sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{(2+\sqrt{2+\sqrt{2+\sqrt{3}}})(2-\sqrt{2+\sqrt{2+\sqrt{3}}})}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{4-(\sqrt{2+\sqrt{2+\sqrt{3}}})^2}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{4-(2+\sqrt{2+\sqrt{3}})}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{2-\sqrt{2+\sqrt{3}}}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{(2+\sqrt{2+\sqrt{3}})\cdot(2-\sqrt{2+\sqrt{3}})}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{4-(\sqrt{2+\sqrt{3}})^2}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{4-(2+\sqrt{3})}$
$\to H=\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}$
$\to H=\sqrt{(2+\sqrt{3})(2-\sqrt{3})}$
$\to H=\sqrt{4-3}$
$\to H=1$
