\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{\frac{1}{2}}\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\sqrt{3}+3}-\sqrt{3-2\sqrt{3}+1}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{\frac{1}{2}}\left(1+\sqrt{3}-\sqrt{3}+1\right)=\frac{1}{\sqrt{2}}.2=\sqrt{2}\)
A = \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3+2.\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
Mà \(\sqrt{3}+1>0;\sqrt{3}-1>\sqrt{1}-1=0\) nên:
\(A=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Đúng ko ta?:3
