Question

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

Answer 1

Lời giải:

Ta có:

\(\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}\) (nhân cả tử và mẫu với $\sqrt{2}$)

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

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

\(=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{4+2\sqrt{3}}{3-1}=2+\sqrt{3}\)