Tính: \(\displaystyle \left( {{{2\sqrt 3 - \sqrt 6 } \over {\sqrt 8 - 2}} - {{\sqrt {216} } \over 3}} \right).{1 \over {\sqrt 6 }} \)

* Hướng dẫn giải

\(\eqalign{
&\left( {{{2\sqrt 3 - \sqrt 6 } \over {\sqrt 8 - 2}} - {{\sqrt {216} } \over 3}} \right).{1 \over {\sqrt 6 }} \cr & =\left( {{{\sqrt 2.\sqrt 2.\sqrt 3 - \sqrt 6 } \over {\sqrt {2^2.2} - 2}} - {{\sqrt {6^2.6} } \over 3}} \right).{1 \over {\sqrt 6 }} \cr & =\left( {{{\sqrt 2.\sqrt 6 - \sqrt 6 } \over {2\sqrt 2 - 2}} - {6.{\sqrt {6} } \over 3}} \right).{1 \over {\sqrt 6 }} \cr 
& = \left[ {{{\sqrt 6 \left( {\sqrt 2 - 1} \right)} \over {2\left( {\sqrt 2 - 1} \right)}} - {{6\sqrt 6 } \over 3}} \right].{1 \over {\sqrt 6 }} \cr 
& = \left( {{{\sqrt 6 } \over 2} - 2\sqrt 6 } \right).{1 \over {\sqrt 6 }}\cr& = \left( {\frac{{\sqrt 6 }}{2} - \frac{{4\sqrt 6 }}{2}} \right).\frac{1}{{\sqrt 6 }} \cr 
& = \left( {{{ - 3} \over 2}\sqrt 6 } \right).{1 \over {\sqrt 6 }} \cr 
& = - {3 \over 2} = - 1,5 \cr} \)