a: \(2\sqrt{2}=\sqrt{8}>\sqrt{7}\)
b: \(3\sqrt{2}=\sqrt{18}>\sqrt{12}=2\sqrt{3}\)
`(a):`
`2\sqrt{2}=\sqrt{4}.\sqrt{2}=\sqrt{8}`
Vì : `\sqrt{8}>\sqrt{7}=>2\sqrt{2}>\sqrt{7}`
`(b):`
`3\sqrt{2}=\sqrt{9}.\sqrt{2}=\sqrt{18}`
`2\sqrt{3}=\sqrt{4}.\sqrt{3}=\sqrt{12}`
Vì : `\sqrt{18}>\sqrt{12}=>3\sqrt{2}>2\sqrt{3}`
a) \(2\sqrt{2}=\sqrt{8}>\sqrt{7}\)
b) \(3\sqrt{2}=\sqrt{18}>2\sqrt{3}=\sqrt{12}\)
a) \(2\sqrt{2}\) và \(\sqrt{7}\)
Ta có: \(2\sqrt{2}=\sqrt{2^2\cdot2}=\sqrt{8}\)
Mà 8 > 7 nên \(2\sqrt{2}>7\)
b) \(3\sqrt{2}\) và \(2\sqrt{3}\)
Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{12}\)
Mà 18 > 12 nên \(3\sqrt{2}>2\sqrt{3}\)
