Rút gọn các biểu thức sau:
a)  \(\left( {\sqrt 8  - 3.\sqrt 2  + \sqrt {10} } \right)\sqrt 2  - \sqrt 5 \)
b)  \(0,2\sqrt {{{\left( { - 10} \right)}^2}.3}  + 2\sqrt {{{\left( {\sqrt 3  - \sqrt 5 } \right)}^2}} \)
c)  \(\left(  - {3 \over 2}.\sqrt 2  + {4 \over 5}.\sqrt {200} } \right):{1 \over 8}\)
d)  \(2\sqrt {{{\left( {\sqrt 2  - 3} \right)}^2}}  + \sqrt {2.{{\left( { - 3} \right)}^2}}  - 5\sqrt {{{\left( { - 1} \right)}^4}} \)
Hướng dẫn làm bài:
a)
\(\eqalign{
& \left( {\sqrt 8 - 3.\sqrt 2 + \sqrt {10} } \right)\sqrt 2 - \sqrt 5 \cr
& = \sqrt {16} - 6 + \sqrt {20} - \sqrt 5 \cr
& = 4 - 6 + 2\sqrt 5 - \sqrt 5 = - 2 + \sqrt 5 \cr} \)                         
b)
\(\eqalign{
& 0,2\sqrt {{{\left( { - 10} \right)}^2}.3} + 2\sqrt {{{\left( {\sqrt 3 - \sqrt 5 } \right)}^2}} \cr
& = 0,2\left| { - 10} \right|\sqrt 3 + 2\left| {\sqrt 3 - \sqrt 5 } \right| \cr
& = 0,2.10.\sqrt 3 + 2\left( {\sqrt 5 - \sqrt 3 } \right) \cr
& = 2\sqrt 3 + 2\sqrt 5 - 2\sqrt 3 = 2\sqrt 5 \cr} \)
Vì \(- 10 < 0;\sqrt 3  < \sqrt 5  \Leftrightarrow \sqrt 3  - \sqrt 5  < 0\)  
c)  
\(\eqalign{
& \left( - {3 \over 2}.\sqrt 2 + {4 \over 5}.\sqrt {200} } \right):{1 \over 8} \cr
& = \left( }} - {3 \over 2}\sqrt 2 + {4 \over 5}\sqrt {{{10}^2}.2} } \right):{1 \over 8} \cr
& = \left( + \sqrt {2.{{\left( { - 3} \right)}^2}} - 5\sqrt {{{\left( { - 1} \right)}^4}} \cr
& = 2\left| {\sqrt 2 - 3} \right| + \left| { - 3} \right|\sqrt 2 - 5\left| { - 1} \right| \cr
& = 2\left( {3 - \sqrt 2 } \right) + 3\sqrt 2 - 5 \cr
& = 6 - 2\sqrt 2 + 3\sqrt 2 - 5 = 1 + \sqrt 2 \cr} \)