Cách 1Đặt $A=\sqrt{2-\sqrt{3} }+\sqrt{2+\sqrt{3} } $.Ta có $A^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{(2-\sqrt{3})(2+\sqrt{3}) }=4+2\sqrt{4-3}=6$.Vậy $A=\sqrt 6.$
Cách 1Đặt $A=\sqrt{2-\sqrt{3} }+\sqrt{2+\sqrt{3} } $.Ta có $A^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{(2-\sqrt{3})(2-\sqrt{3}) }=4+2\sqrt{4-3}=6$.Vậy $A=\sqrt 6.$
Cách 1Đặt $A=\sqrt{2-\sqrt{3} }+\sqrt{2+\sqrt{3} } $.Ta có $A^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{(2-\sqrt{3})(2
+\sqrt{3}) }=4+2\sqrt{4-3}=6$.Vậy $A=\sqrt 6.$