Lời giải:
$\sqrt{17}+\sqrt{10}> \sqrt{16}+\sqrt{9}=4+3=7$
\(\sqrt[]{17}+\sqrt[]{10}\Rightarrow\left(\sqrt[]{17}+\sqrt[]{10}\right)^2=17+10+2\sqrt[]{70}=27+2\sqrt[]{70}< 27+2\sqrt[]{100}=47\)
mà \(7^2=49>47\)
\(\Rightarrow\sqrt[]{17}+\sqrt[]{10}< 7\)
√17 = 8,5
√10 = 5
=> 8,5 > 7 > 5
=> √17 > 7 > √10