Khử mẫu của biểu thức lấy căn:

    a). \(\sqrt {\frac{1}{{600}}} ;\,\,\,\sqrt {\frac{{11}}{{540}}} ;\,\,\,\sqrt {\frac{3}{{50}}} ;\,\,\,\sqrt {\frac{5}{{98}}} ;\,\,\,\,\sqrt {\frac{{{{\left( {1 - \sqrt 3 } \right)}^2}}}{{27}}} .\)       b). \(ab\sqrt {\frac{a}{b}} ;\) \(\frac{a}{b}\sqrt {\frac{b}{a}} ;\) \(\sqrt {\frac{1}{b} + \frac{1}{{{b^2}}}} ;\) \(\sqrt {\frac{{9{a^3}}}{{36b}}} ;\) \(3xy\sqrt {\frac{2}{{xy}}} .\)

(Giả thiết các biểu thức có nghĩa).

a). \[\sqrt {\frac{1}{{600}}}  = \sqrt {\frac{{600}}{{{{600}^2}}}}  = \frac{{\sqrt {6.100} }}{{600}} = \frac{{10\sqrt 6 }}{{600}} = \frac{{\sqrt 6 }}{{60}};\]

\(\sqrt {\frac{{11}}{{540}}}  = \frac{{\sqrt {11} }}{{\sqrt {9.4.15} }} = \frac{{\sqrt {11} }}{{6\sqrt {15} }} = \frac{{\sqrt {11} .\sqrt {15} }}{{6.15}} = \frac{{\sqrt {165} }}{{90}};\)

\(\sqrt {\frac{3}{{50}}}  = \sqrt {\frac{{3.50}}{{{{50}^2}}}}  = \frac{{\sqrt {6.25} }}{{50}} = \frac{{5\sqrt 6 }}{{50}} = \frac{{\sqrt 6 }}{{10}};\)

\(\sqrt {\frac{5}{{98}}}  = \frac{{\sqrt 5 }}{{\sqrt {49.2} }} = \frac{{\sqrt 5 }}{{7\sqrt 2 }} = \frac{{\sqrt {10} }}{{14}};\)

\(\sqrt {\frac{{{{\left( {1 - \sqrt 3 } \right)}^2}}}{{27}}}  = \frac{{\sqrt 3  - 1}}{3}.\frac{{\sqrt 3 }}{3} = \frac{{\sqrt 3 \left( {\sqrt 3  - 1} \right)}}{9}\)

b). \(ab\sqrt {\frac{a}{b}}  = ab\sqrt {\frac{{ab}}{{{b^2}}}}  = \frac{{ab}}{{\left| b \right|}}\sqrt {ab}  = \left\{ \begin{array}{l}a\sqrt {ab} \,\,{\rm{neu}}\,\,\,b > 0\\ - a\sqrt {ab} \,neu\,\,b < 0\end{array} \right.;\)

\(\frac{a}{b}\sqrt {\frac{b}{a}}  = \frac{a}{b}\sqrt {\frac{{ab}}{{{a^2}}}}  =  = \frac{a}{{b\left| a \right|}}\sqrt {ab}  = \left\{ \begin{array}{l}\frac{{\sqrt {ab} }}{b}\,neu\,a > 0\\ - \frac{{\sqrt {ab} }}{b}\,neu\,\,a < 0\end{array} \right.\)

\(\sqrt {\frac{1}{b} + \frac{1}{{{b^2}}}}  = \sqrt {\frac{{1 + b}}{{{b^2}}}}  = \frac{{\sqrt {1 + b} }}{{\left| b \right|}};\)

\(\sqrt {\frac{{9{a^3}}}{{36b}}}  = \frac{{\sqrt {{a^3}b} }}{{2\left| b \right|}} = \frac{{\left| a \right|\sqrt {ab} }}{{2\left| b \right|}} = \frac{{a\sqrt {ab} }}{{2b}}\)(\(ab \ge 0;b \ne 0).\)

\(3xy\sqrt {\frac{2}{{xy}}}  = 3xy.\frac{{\sqrt {2xy} }}{{xy}} = 3\sqrt {2xy} \) (vì \(xy \ge 0).\)