a. Ta co: \(2\sqrt{31}=\sqrt{2^2\cdot31}=\sqrt{124}\)
\(10=\sqrt{10}\)
\(Vi\sqrt{10}< \sqrt{124}hay10< 2\sqrt{31}\)
c.. \(6+2\sqrt{2}=6+\sqrt{8}=\sqrt{6}+\sqrt{8}=\sqrt{14}\)(1)
\(9=\sqrt{3}\)(2)
Ta co (1)>(2) hay \(\sqrt{14}>\sqrt{3}\) nen \(6+2\sqrt{2}>9\)
f. \(\sqrt{11}-\sqrt{3}=\sqrt{9}=3\)
Vì 3>2 nên \(\sqrt{11}-\sqrt{3}>2\)