Question

Answer 1

1: Đặt \(B=\sqrt{3-\sqrt2}+\sqrt{3+\sqrt2}\)

=>\(B^2=3-\sqrt2+3+\sqrt2+2\cdot\sqrt{9-2}=6+2\sqrt7\)

=>\(B=\sqrt{6+2\sqrt7}\)

Ta có: \(A=\frac{\sqrt{3-\sqrt2}+\sqrt{3+\sqrt2}}{\sqrt{3+\sqrt7}}+\sqrt{5-2\sqrt6}\)

\(=\sqrt{\frac{6+2\sqrt7}{3+\sqrt7}}+\sqrt{\left(\sqrt3-\sqrt2\right)^2}=\sqrt2+\sqrt3-\sqrt2=\sqrt3\)