a) Ta có:\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)\(\eqalign{
& = \sqrt {{5^2}.{1 \over 5}} + \sqrt {{{\left( {{1 \over 2}} \right)}^2}.20} + \sqrt 5 \cr
& = \sqrt {25.{1 \over 5}} + \sqrt {{1 \over 4}.20} + \sqrt 5 \cr
& = \sqrt {{{25} \over 5}} + \sqrt {{{20} \over 4}} + \sqrt 5 \cr
& = \sqrt 5 + \sqrt 5 + \sqrt 5 \cr
& = \left( {1 + 1 + 1} \right)\sqrt 5 = 3\sqrt 5 \cr} \)
b) Ta có: \(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\)\(\eqalign{
& = \sqrt {{1 \over 2}} + \sqrt {{9 \over 2}} + \sqrt {{{25} \over 2}} \cr
& = \sqrt {{1 \over 2}} + \sqrt {9.{1 \over 2}} + \sqrt {25.{1 \over 2}} \cr
& = \sqrt {{1 \over 2}} + \sqrt {3^2.{1 \over 2}} + \sqrt {5^2.{1 \over 2}} \cr
& = \sqrt {{1 \over 2}} + 3\sqrt {{1 \over 2}} + 5\sqrt {{1 \over 2}} \cr
& = \left( {1 + 3 + 5} \right).\sqrt {{1 \over 2}} \cr
& = 9\sqrt {{1 \over 2}} = 9{1 \over {\sqrt 2 }} \cr
& = 9.{{\sqrt 2 } \over 2} = {{9\sqrt 2 } \over 2} \cr} \)
c) Ta có:\(\eqalign{
& \sqrt {20} - \sqrt {45} + 3\sqrt {18} + \sqrt {72} \cr
& = \sqrt {4.5} - \sqrt {9.5} + 3\sqrt {9.2} + \sqrt {36.2} \cr
& = \sqrt {{2^2}.5} - \sqrt {{3^2}.5} + 3\sqrt {{3^2}.2} + \sqrt {{6^2}.2} \cr
& = 2\sqrt 5 - 3\sqrt 5 + 3.3\sqrt 2 + 6\sqrt 2 \cr
& = 2\sqrt 5 - 3\sqrt 5 + 9\sqrt 2 + 6\sqrt 2 \cr
& = \left( {2\sqrt 5 - 3\sqrt 5 } \right) + \left( {9\sqrt 2 + 6\sqrt 2 } \right) \cr
& = \left( {2 - 3} \right)\sqrt 5 + \left( {9 + 6} \right)\sqrt 2 \cr
& = - \sqrt 5 + 15\sqrt 2 = 15\sqrt 2 - \sqrt 5 \cr} \)
d) Ta có:\(\eqalign{
& 0,1\sqrt {200} + 2\sqrt {0,08} + 0,4.\sqrt {50} \cr
& = 0,1\sqrt {100.2} + 2\sqrt {0,04.2} + 0,4\sqrt {25.2} \cr
& = 0,1\sqrt {10^2.2} + 2\sqrt {0,2^2.2} + 0,4\sqrt {5^2.2} \cr
& = 0,1.10\sqrt 2 + 2.0,2\sqrt 2 + 0,4.5\sqrt 2 \cr
& = 1\sqrt 2 + 0,4\sqrt 2 + 2\sqrt 2 \cr
& = \left( {1 + 0,4 + 2} \right)\sqrt 2 = 3,4\sqrt 2 \cr} \)
