Question

So sánh:

a) \(2\sqrt{3}\) và \(3\sqrt{2}\);                      b) \(7\sqrt{\dfrac{3}{7}}\) và \(\sqrt{2}.\sqrt{11}\);                    c) \(\dfrac{2}{\sqrt{5}}\) và \(\dfrac{6}{\sqrt{10}}\).

Answer 1

a. Ta có: \(2\sqrt 3  = \sqrt {12} ;\,\,3\sqrt 2  = \sqrt {18} \).

Do \(12 < 18\) nên \(\sqrt {12}  < \sqrt {18} \) hay \(2\sqrt 3  < 3\sqrt 2 \).

b. Ta có: \(7\sqrt {\frac{3}{7}}  = \sqrt {21} ;\sqrt 2 .\sqrt {11}  = \sqrt {22} \).

Do \(21 < 22\) nên \(\sqrt {21}  < \sqrt {22} \) hay \(7\sqrt {\frac{3}{7}}  < \sqrt 2 .\sqrt {11} \).

c. Ta có: \(\frac{2}{{\sqrt 5 }} = \frac{{\sqrt 4 }}{{\sqrt 5 }} = \sqrt {\frac{4}{5}} ;\frac{6}{{\sqrt {10} }} = \frac{{\sqrt {36} }}{{\sqrt {10} }} = \sqrt {\frac{{36}}{{10}}}  = \sqrt {\frac{{18}}{5}} \).

Do \(\frac{4}{5} < \frac{{18}}{5}\) nên \(\sqrt {\frac{4}{5}}  < \sqrt {\frac{{18}}{5}} \) hay \(\frac{2}{{\sqrt 5 }} < \frac{6}{{\sqrt {10} }}\).