Trục căn thức ở mẫu : a) \(\dfrac{5}{{3\sqrt 7 }}\); b) \(\dfrac{4}{{\sqrt 5 - 3}}\); c) \(\dfrac{a}{{\sqrt a + 1}}\).
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Ta có:
\(\begin{array}{l}a)\;\;\dfrac{5}{{3\sqrt 7 }} = \dfrac{{5.\sqrt 7 }}{{3.7}} = \dfrac{{5\sqrt 7 }}{{21}}. & & \\b)\;\;\dfrac{4}{{\sqrt 5 - 3}} = \dfrac{{4\left( {\sqrt 5 + 3} \right)}}{{\left( {\sqrt 5 - 3} \right)\left( {\sqrt 5 + 3} \right)}} = \dfrac{{4\left( {\sqrt 5 + 3} \right)}}{{5 - 9}} = - \dfrac{{4\left( {\sqrt 5 + 3} \right)}}{4} = \sqrt 5 + 3.\\c)\;\;\dfrac{a}{{\sqrt a + 1}} = \dfrac{{a\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}} = \dfrac{{a\left( {\sqrt a - 1} \right)}}{{a - 1}}.\end{array}\)