Trục căn thức ở mẫu các biểu thức:
a) \(\frac{{5\sqrt 2 }}{{\sqrt {15} }}\)
b) \( - \frac{{2\sqrt 5 }}{{\sqrt {18} }}\)
c) \(\frac{{6a}}{{\sqrt {2a{b^2}} }}(a > 0;b > 0)\)
Dựa vào: \(\frac{{\sqrt a }}{{\sqrt b }} = \frac{{\sqrt a .\sqrt b }}{{{{\left( {\sqrt b } \right)}^2}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
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\(\sqrt {\frac{a}{b}} = \sqrt {\frac{{ab}}{{{b^2}}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
a) \(\frac{{5\sqrt 2 }}{{\sqrt {15} }} = \frac{{5\sqrt 2 .\sqrt {15} }}{{\sqrt {15} .\sqrt {15} }} = \frac{{5\sqrt {30} }}{{15}} = \frac{{\sqrt {30} }}{3}\)
b) \( - \frac{{2\sqrt 5 }}{{\sqrt {18} }} = - \frac{{2\sqrt 5 }}{{3\sqrt 2 }} = - \frac{{2\sqrt 5 .\sqrt 2 }}{{3\sqrt 2 .\sqrt 2 }} = \frac{{\sqrt {10} }}{3}\)
c) \(\frac{{6a}}{{\sqrt {2a{b^2}} }}\)
\(= \frac{{6a}}{{b\sqrt {2a} }} = \frac{{6a.\sqrt {2a} }}{{b\sqrt {2a} .\sqrt {2a} }} = \frac{{6a\sqrt {2a} }}{{2ab}} = \frac{{3\sqrt {2a} }}{b}(a > 0;b > 0)\)