1 bài tập Khử mẫu của biểu thức lấy căn (có lời giải)

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 $xy \ge 0).$