Question

Tính \(A=\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}\)

Answer 1

\(A=\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}=\dfrac{\sqrt{2}\left(\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}\right)}{\sqrt{2}}=\dfrac{\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{11}-1\right)^2}-\sqrt{\left(\sqrt{11}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{11}-1-\sqrt{11}-1}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)

Answer 2

\(A=\sqrt{\left(\sqrt{\dfrac{11}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\dfrac{11}{2}+\sqrt{\dfrac{1}{2}}\right)^2}\\ A=\sqrt{\dfrac{11}{2}}-\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{11}{2}}-\sqrt{\dfrac{1}{2}}\\ A=-2\sqrt{\dfrac{1}{2}}=-\dfrac{2\sqrt{2}}{2}=-\sqrt{2}\)

Answer 3

\(A=\dfrac{\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{11}-1\right)^2}-\sqrt{\left(\sqrt{11}+1\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11}-1-\sqrt{11}-1}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)