a: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)
\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
mà 105<120
nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b: \(\sqrt{8}+\sqrt{2}=\dfrac{6}{\sqrt{8}-\sqrt{2}}\)
\(3+\sqrt{3}=\dfrac{6}{3-\sqrt{3}}\)
mà căn 8<3; -căn 2>-căn 3
nên \(\sqrt{8}+\sqrt{2}< 3+\sqrt{3}\)