Question

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

Answer 1

Đặt: \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

\(A^2=2+\sqrt{3}+2-\sqrt{3}+2\)

\(A^2=6\)

\(A=\sqrt{6}\)

Answer 2

Cách khác :D \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)