Trục căn thức ở mẫu các biểu thức sau:
a) \(\frac{{\sqrt 6 + 2}}{{\sqrt 6 - 2}}\)
b) \(\frac{1}{{\sqrt 2 (\sqrt 5 - 1)}}\)
c) \(\frac{{x - 1}}{{2\sqrt x - \sqrt {x + 3} }}(x \ge 0,x \ne 1)\)
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Dựa vào: \(\frac{{\sqrt a }}{{\sqrt b }} = \frac{{\sqrt a .\sqrt b }}{{{{\left( {\sqrt b } \right)}^2}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
\(\sqrt {\frac{a}{b}} = \sqrt {\frac{{ab}}{{{b^2}}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
a) \(\frac{{\sqrt 6 + 2}}{{\sqrt 6 - 2}} = \frac{{{{\left( {\sqrt 6 + 2} \right)}^2}}}{{\left( {\sqrt 6 - 2} \right)\left( {\sqrt 6 + 2} \right)}} = \frac{{6 + 4\sqrt 6 + 4}}{{6 - 4}} = 5 + 2\sqrt 6 .\)
b) \(\frac{1}{{\sqrt 2 (\sqrt 5 - 1)}} = \frac{{\sqrt 2 (\sqrt 5 + 1)}}{{{{\left( {\sqrt 2 } \right)}^2}(\sqrt 5 - 1)(\sqrt 5 + 1)}} = \frac{{\sqrt {10} + \sqrt 2 }}{{2.4}} = \frac{{\sqrt {10} + \sqrt 2 }}{8}.\)
c) \(\frac{{x - 1}}{{2\sqrt x - \sqrt {x + 3} }} = \frac{{\left( {x - 1} \right)(2\sqrt x - \sqrt {x + 3} )}}{{4x - (x + 3)}} = \frac{{2\sqrt x + \sqrt {x + 3} }}{3}.\)