a) Ta có :\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\) =\(\sqrt{\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
Tương tự : \(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\) = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
=>\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\)+\(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)+\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)= \(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)=\(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{3-2}\)=10
bạn nhìn lại đề xem có sai gì không bạn
b) Ta có : \(\sqrt{5}\)-\(\sqrt{3-\sqrt{29-6\sqrt{20}}}\)= \(\sqrt{5}-\sqrt{3-\sqrt{29-2\sqrt{180}}}\)=\(\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)^2}\)=\(\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)=\(\sqrt{5}-\sqrt{6-2\sqrt{5}}\)=\(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)=\(\sqrt{5}-\sqrt{5}+1\)=1