1 )A =\(\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)
B =\(\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}\)
=\(\left(1+\sqrt{180}\right)-\sqrt{21}>0.\)
\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)
2 ) C =\(\left(\sqrt{5}+\sqrt{10}+1\right)^2\)
=\(5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)
=\(26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(35\right)^2\)
\(V\text{ậ}y\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)
3 )\(\left(\dfrac{15-2\sqrt{10}}{3}\right)^2=\dfrac{225-60\sqrt{10}+40}{9}\)
\(=\dfrac{265-60\sqrt{10}}{9}\)
\(=\dfrac{265}{9}-\dfrac{20\sqrt{10}}{3}< 15.\)
Vậy nên \(\dfrac{15-2\sqrt{10}}{3}< \sqrt{15}.\)