a)
<=> \(\dfrac{7}{4\cdot\sqrt{3}}và\dfrac{9}{4\cdot\sqrt{5}}\)
<=> \(\dfrac{7\cdot\sqrt{5}}{4\cdot\sqrt{15}}và\dfrac{9\cdot\sqrt{3}}{4\cdot\sqrt{15}}\)
<=>\(\sqrt{245}và\sqrt{243}\)
<=> \(\sqrt{245}>\sqrt{243}\)
=> \(\dfrac{7}{2}\cdot\sqrt{\dfrac{1}{12}}=\dfrac{9}{4}\cdot\sqrt{\dfrac{1}{5}}\)
a)
\(\dfrac{7}{2}\sqrt{\dfrac{1}{12}}=\dfrac{7}{2}\sqrt{\dfrac{12}{12^2}}=\dfrac{7}{2}.\dfrac{\sqrt{12}}{\sqrt{12^2}}=\dfrac{7}{2}.\dfrac{\sqrt{3.4}}{12}=\dfrac{7.2.\sqrt{3}}{2.12}=\dfrac{7\sqrt{3}}{12}=\dfrac{7\sqrt{3}.5}{12.5}=\dfrac{35\sqrt{3}}{60}\)
\(\dfrac{9}{4}\sqrt{\dfrac{1}{5}}=\dfrac{9}{4}\sqrt{\dfrac{5}{5^2}}=\dfrac{9}{4}.\dfrac{\sqrt{5}}{\sqrt{5^2}}=\dfrac{9.\sqrt{5}}{4.5}=\dfrac{9\sqrt{5}}{20}=\dfrac{9\sqrt{5}.3}{20.3}=\dfrac{27\sqrt{5}}{60}\)Ta có \(3675>3645\Leftrightarrow\sqrt{3675}>\sqrt{3645}\Leftrightarrow\sqrt{1225.3}>\sqrt{729.5}\Leftrightarrow35\sqrt{3}>27\sqrt{5}\Leftrightarrow\dfrac{35\sqrt{3}}{60}>\dfrac{27\sqrt{5}}{60}\)
Vậy \(\dfrac{7}{2}\sqrt{\dfrac{1}{12}}>\dfrac{9}{4}\sqrt{\dfrac{1}{5}}\)
b)
\(\sqrt{\dfrac{4}{27}}=\sqrt{\dfrac{4.3}{27.3}}=\dfrac{\sqrt{4.3}}{\sqrt{81}}=\dfrac{2\sqrt{3}}{9}=\dfrac{2\sqrt{3}.26}{9.26}=\dfrac{52\sqrt{3}}{234}\)
\(\sqrt{\dfrac{3}{26}}=\sqrt{\dfrac{3.26}{26^2}}=\dfrac{\sqrt{3.26}}{\sqrt{26^2}}=\dfrac{\sqrt{78}}{26}=\dfrac{9.\sqrt{78}}{26.9}=\dfrac{9\sqrt{78}}{234}\)
Ta có \(8112>6318\Leftrightarrow\sqrt{8112}>\sqrt{6318}\Leftrightarrow\sqrt{2704.3}>\sqrt{81.78}\Leftrightarrow52\sqrt{3}>9\sqrt{78}\Leftrightarrow\dfrac{52\sqrt{3}}{234}>\dfrac{9\sqrt{78}}{234}\)
Vậy \(\sqrt{\dfrac{4}{27}}>\sqrt{\dfrac{3}{26}}\)
