Chọn khẳng định đúng?

Ta có $\left( {\dfrac{{\sqrt {14}  - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15}  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{a\left( {\sqrt 7  - \sqrt 5 } \right)}}$$= \left( {\dfrac{{\sqrt 7 .\sqrt 2  - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 .\sqrt 3  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right).a\left( {\sqrt 7  - \sqrt 5 } \right)$$= \left( {\dfrac{{ - \sqrt 7 \left( {1 - \sqrt 2 } \right)}}{{1 - \sqrt 2 }} - \dfrac{{\sqrt 5 \left( {\sqrt 3  - 1} \right)}}{{\sqrt 3  - 1}}} \right).a\left( {\sqrt 7  - \sqrt 5 } \right)$

$= \left[ {\dfrac{{\sqrt 6 \left( {\sqrt 2  - 1} \right)}}{{2\left( {\sqrt 2  - 1} \right)}} - 2\sqrt 6 } \right].\left( { - \dfrac{a}{{\sqrt 6 }}} \right)$$= \left( { - \sqrt 7  - \sqrt 5 } \right).a\left( {\sqrt 7  - \sqrt 5 } \right) =  - a.\left( {\sqrt 7  + \sqrt 5 } \right)\left( {\sqrt 7  - \sqrt 5 } \right) =  - 2a.$