a) Ta có:
\(3\sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}\).
Vì \( 27>12 \Leftrightarrow \sqrt{27} > \sqrt{12}\)
\(\Leftrightarrow 3\sqrt{3} >\sqrt{12}\).
Vậy: \(3\sqrt{3}>\sqrt{12}\).
b) Ta có:
\(7=\sqrt{7^2}=\sqrt{49}\).
\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{9.5}=\sqrt{45}\).
Vì \(49> 45 \Leftrightarrow \sqrt {49}> \sqrt {45} \Leftrightarrow 7 >3\sqrt 5\).
Vậy: \(7>3\sqrt{5}\).
c) Ta có:
\(\dfrac{1}{3}\sqrt{51}= \sqrt {{\left(\dfrac{1}{3} \right)}^2.51 } = \sqrt {\dfrac{1}{9}.51} = \sqrt {\dfrac{51}{9}} \)
\(= \sqrt {\dfrac{3.17}{3.3}} = \sqrt {\dfrac{17}{3}} \).
\(\dfrac{1}{5}\sqrt{150}= \sqrt {{\left(\dfrac{1}{5} \right)}^2.150 } = \sqrt {\dfrac{1}{25}.150} = \sqrt {\dfrac{150}{25}} \)
\(= \sqrt {\dfrac{6.25}{25}} = \sqrt {6}=\sqrt{\dfrac{18}{3}} \).
Vì \( \dfrac{17}{3} <\dfrac{18}{3} \Leftrightarrow \sqrt{\dfrac{17}{3}} < \sqrt{\dfrac{18}{3}}\)
\(\Leftrightarrow \dfrac{1}{3}\sqrt{51} <\dfrac{1}{5}\sqrt{150}\).
Vậy: \( \dfrac{1}{3}\sqrt{51} <\dfrac{1}{5}\sqrt{150}\).
d) Ta có:
\(\dfrac{1}{2}\sqrt{6}= \sqrt {{\left(\dfrac{1}{2} \right)}^2.6 } = \sqrt {\dfrac{1}{4}.6} = \sqrt {\dfrac{6}{4}} = \sqrt {\dfrac{2.3}{2.2}} \)
\(= \sqrt {\dfrac{3}{2}} \).
\(6\sqrt{\dfrac{1}{2}}=\sqrt{6^2.\dfrac{1}{2}}=\sqrt{36.\dfrac{1}{2}}=\sqrt{\dfrac{36}{2}}\).
Vì \( \dfrac{3}{2}<\dfrac{36}{2} \Leftrightarrow \sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}\)
\(\Leftrightarrow \dfrac{1}{2}\sqrt{6} <6\sqrt{\dfrac{1}{2}}\).
Vậy: \(\dfrac{1}{2}\sqrt{6}<6\sqrt{\dfrac{1}{2}}\).
