a) Ta có:
\(2\sqrt{6}=\sqrt{2^2\cdot6}=\sqrt{4\cdot6}=\sqrt{24}\)
\(3\sqrt{3}=\sqrt{3^2\cdot3}=\sqrt{9\cdot3}=\sqrt[]{27}\)
Mà: \(24< 27\)
\(\Rightarrow\sqrt{24}< \sqrt{27}\)
\(\Rightarrow2\sqrt{6}< 3\sqrt{3}\)
Vậy: \(3\sqrt{3}\) là số lớn hơn
b) Ta có:
\(\dfrac{2}{5}\sqrt{6}=\sqrt{\left(\dfrac{2}{5}\right)^2\cdot6}=\sqrt{\dfrac{4\cdot6}{25}}=\sqrt{\dfrac{24}{25}}\)
\(\dfrac{7}{4}\sqrt{\dfrac{1}{3}}=\sqrt{\left(\dfrac{7}{4}\right)^2\cdot\dfrac{1}{3}}=\sqrt{\dfrac{49}{16\cdot3}}=\sqrt{\dfrac{49}{48}}\)
Mà: \(\dfrac{24}{25}< 1< \dfrac{49}{48}\)
\(\Rightarrow\dfrac{24}{25}< \dfrac{49}{48}\)
\(\Rightarrow\sqrt{\dfrac{24}{25}}< \sqrt{\dfrac{49}{48}}\)
\(\Rightarrow\dfrac{2}{5}\sqrt{6}< \dfrac{7}{4}\sqrt{\dfrac{1}{3}}\)
Vậy: \(\dfrac{7}{4}\sqrt{\dfrac{1}{3}}\) là số lớn hơn
a: \(2\sqrt{6}=\sqrt{24}\)
\(3\sqrt{3}=\sqrt{27}\)
mà 24<27
nên \(2\sqrt{6}< 3\sqrt{3}\)
=>3căn 3 là số lớn hơn
b: \(\dfrac{2}{5}\sqrt{6}=\sqrt{\dfrac{4}{25}\cdot6}=\sqrt{\dfrac{24}{25}}\)<căn 25/25=1
\(\dfrac{7}{4}\sqrt{\dfrac{1}{3}}=\sqrt{\dfrac{49}{16}\cdot\dfrac{1}{3}}=\sqrt{\dfrac{49}{48}}>1\)
=>\(\sqrt{\dfrac{24}{25}}< 1< \sqrt{\dfrac{49}{48}}\)
=>\(\dfrac{2}{5}\sqrt{6}< \dfrac{7}{4}\sqrt{\dfrac{1}{3}}\)
=>\(\dfrac{7}{4}\sqrt{\dfrac{1}{3}}\) là số lớn hơn
