Trục căn thức ở mẫu các biểu thức sau:
a) \(\frac{4}{{\sqrt {13} - 3}}\)
b) \(\frac{{10}}{{5 + 2\sqrt 5 }}\)
c) \(\frac{{\sqrt a - \sqrt b }}{{\sqrt a + \sqrt b }}\) với a > 0; b > 0, \(a \ne b\).
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Dựa vào VD3 trang 53 làm tương tự.
a) \(\frac{4}{{\sqrt {13} - 3}}\)\( = \frac{{4.\left( {\sqrt {13} + 3} \right)}}{{\left( {\sqrt {13} - 3} \right)\left( {\sqrt {13} + 3} \right)}}\)\( = \frac{{4.\left( {\sqrt {13} + 3} \right)}}{{{{\left( {\sqrt {13} } \right)}^2} - {3^2}}}\)\( = \frac{{4.\left( {\sqrt {13} + 3} \right)}}{{13 - 9}}\)\( = \left( {\sqrt {13} + 3} \right)\)
b) \(\frac{{10}}{{5 + 2\sqrt 5 }}\)\( = \frac{{10.\left( {5 - 2\sqrt 5 } \right)}}{{\left( {5 + 2\sqrt 5 } \right)\left( {5 - 2\sqrt 5 } \right)}}\)\( = \frac{{10.\left( {5 - 2\sqrt 5 } \right)}}{{{5^2} - {{\left( {2\sqrt 5 } \right)}^2}}}\)\( = \frac{{10.\left( {5 - 2\sqrt 5 } \right)}}{5}\)\( = 2\left( {5 - 2\sqrt 5 } \right)\)
c) \(\frac{{\sqrt a - \sqrt b }}{{\sqrt a + \sqrt b }}\)\( = \frac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\)\( = \frac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{{{\left( {\sqrt a } \right)}^2} - {{\left( {\sqrt b } \right)}^2}}}\)\( = \frac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{a - b}}\) với a > 0; b > 0, \(a \ne b\).