Question

Tính:

a)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)

b)\(\sqrt{6.5+\sqrt{12}}+\sqrt{6.5-\sqrt{12}}+2\sqrt{6}\)

Answer 1

a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{\frac{6+2\sqrt{5}}{2}}-\sqrt{\frac{6-2\sqrt{5}}{2}}-\sqrt{2}\)

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

    \(=\frac{\left|\sqrt{3}+\sqrt{2}\right|}{\sqrt{2}}-\frac{\left|\sqrt{3}-\sqrt{2}\right|}{\sqrt{2}}-\sqrt{2}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\frac{2}{\sqrt{2}}\)

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

b, Tương tự