a) Ta có:
\(2\sqrt{5}=\sqrt{2^2\cdot5}=\sqrt{4\cdot5}=\sqrt{20}\)
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(5=\sqrt{5^2}=\sqrt{25}\)
Mà:
\(18< 20< 23< 25\)
\(\Rightarrow\sqrt{18}< \sqrt{20}< \sqrt{23}< \sqrt{25}\)
\(\Rightarrow3\sqrt{2}< 2\sqrt{5}< \sqrt{23}< 5\)
b) Ta có:
\(5\sqrt{2}=\sqrt{5^2\cdot2}=\sqrt{25\cdot2}=\sqrt{50}\)
\(2\sqrt{13}=\sqrt{2^2\cdot13}=\sqrt{4\cdot13}=\sqrt{52}\)
\(4\sqrt{3}=\sqrt{4^2\cdot3}=\sqrt{16\cdot3}=\sqrt{48}\)
Mà:
\(52>50>48>47\)
\(\Rightarrow\sqrt{52}>\sqrt{50}>\sqrt{48}>\sqrt{47}\)
\(\Rightarrow2\sqrt{13}>5\sqrt{2}>4\sqrt{3}>\sqrt{47}\)
a: \(2\sqrt{5}=\sqrt{20};3\sqrt{2}=\sqrt{18};5=\sqrt{25}\)
18<20<23<25
=>\(\sqrt{18}< \sqrt{20}< \sqrt{23}< \sqrt{25}\)
=>\(3\sqrt{2}< 2\sqrt{5}< \sqrt{23}< 5\)
b: \(5\sqrt{2}=\sqrt{50};2\sqrt{13}=\sqrt{52};4\sqrt{3}=\sqrt{48}\)
52>50>48>47
=>\(\sqrt{52}>\sqrt{50}>\sqrt{48}>\sqrt{47}\)
=>\(2\sqrt{13}>5\sqrt{2}>4\sqrt{3}>\sqrt{47}\)
