1 bài tập Trục căn ở mẫu (có lời giải)

a) \(\frac{5}{{\sqrt {10} }} = \frac{{5\sqrt {10} }}{{10}} = \frac{{\sqrt {10} }}{2};\)    \(\frac{5}{{2\sqrt 5 }} = \frac{{{{\left( {\sqrt 5 } \right)}^2}}}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{2};\)

\(\frac{1}{{3\sqrt {20} }} = \frac{1}{{3\sqrt {4.5} }} = \frac{1}{{6\sqrt 5 }} = \frac{{\sqrt 5 }}{{30}};\)

\(\frac{{2\sqrt 2  + 2}}{{5\sqrt 2 }} = \frac{{\sqrt 2 \left( {2 + \sqrt 2 } \right)}}{{5\sqrt 2 }} = \frac{{2 + \sqrt 2 }}{5};\)

\(\frac{{y + b\sqrt y }}{{b\sqrt y }} = \frac{{\sqrt y \left( {\sqrt y  + b} \right)}}{{b\sqrt y }} = \frac{{\sqrt y  + b}}{b}.\)

b)  \(\frac{3}{{\sqrt 3  + 1}} = \frac{{3\left( {\sqrt 3  - 1} \right)}}{{3 - 1}} = \frac{{3\left( {\sqrt 3  - 1} \right)}}{2};\)

\(\frac{2}{{\sqrt 3  - 1}} = \frac{{2\left( {\sqrt 3  + 1} \right)}}{{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}} = \sqrt 3  + 1;\)

\(\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} = \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} = \frac{{7 + 4\sqrt 3 }}{{4 - 3}} = 7 + 4\sqrt 3 ;\)

\(\frac{b}{{3 + \sqrt b }} = \frac{{b\left( {3 - \sqrt b } \right)}}{{9 - b}};\) \(\frac{p}{{2\sqrt p  - 1}} = \frac{{p\left( {2\sqrt p  + 1} \right)}}{{4p - 1}} = \frac{{2p\sqrt p  + p}}{{4p - 1}};\)

c) \(\frac{2}{{\sqrt 6  - \sqrt 5 }} = 2\left( {\sqrt 6  + \sqrt 5 } \right);\)   

 \(\frac{3}{{\sqrt {10}  + \sqrt 7 }} = \frac{{3\left( {\sqrt {10}  - \sqrt 7 } \right)}}{{10 - 7}} = \sqrt {10}  - \sqrt 7 ;\)

\(\frac{1}{{\sqrt x  - \sqrt y }} = \frac{{\sqrt x  + \sqrt y }}{{x - y}};\)     

\(\frac{{2ab}}{{\sqrt a  - \sqrt b }} = \frac{{2ab\left( {\sqrt a  + \sqrt b } \right)}}{{a - b}}.\)