Tính: \(\displaystyle \left( {{{\sqrt {14} - \sqrt 7 } \over {1 - \sqrt 2 }} + {{\sqrt {15} - \sqrt 5 } \over {1 - \sqrt 3 }}} \right):{ 1 \over {\sqrt 7 - \sqrt 5 }} \)

* Hướng dẫn giải

\(\eqalign{
& \left( {{{\sqrt {14} - \sqrt 7 } \over {1 - \sqrt 2 }} + {{\sqrt {15} - \sqrt 5 } \over {1 - \sqrt 3 }}} \right):{1 \over {\sqrt 7 - \sqrt 5 }} \cr &= \left( {{{\sqrt {7}.\sqrt 2 - \sqrt 7 } \over {1 - \sqrt 2 }} + {{\sqrt {5}.\sqrt 3 - \sqrt 5 } \over {1 - \sqrt 3 }}} \right):{1 \over {\sqrt 7 - \sqrt 5 }} \cr
& = \left[ {{{\sqrt 7 \left( {\sqrt 2 - 1} \right)} \over {1 - \sqrt 2 }} + {{\sqrt {5 }\left( {\sqrt 3 - 1} \right)} \over {1 - \sqrt 3 }}} \right]:{1 \over {\sqrt 7 - \sqrt 5 }} \cr 
& = \left( { - \sqrt 7 - \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right) \cr 
& = - \left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right) \cr 
& = - \left( {7 - 5} \right) = - 2 \cr} \)