Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2\sqrt {50} - 3\sqrt {\dfrac{1}{2}} + 4\sqrt {32} - 5\sqrt {72} \\
= 2\sqrt {25.2} - 3.\dfrac{{\sqrt 1 }}{{\sqrt 2 }} + 4.\sqrt {16.2} - 5\sqrt {36.2} \\
= 2\sqrt {{5^2}.2} - 3.\dfrac{1}{{\sqrt 2 }} + 4.\sqrt {{4^2}.2} - 5.\sqrt {{6^2}.2} \\
= 2.5.\sqrt 2 - \dfrac{{3.\sqrt 2 }}{2} + 4.4\sqrt 2 - 5.6\sqrt 2 \\
= 10\sqrt 2 - \dfrac{3}{2}.\sqrt 2 + 16\sqrt 2 - 30\sqrt 2 \\
= - \dfrac{{11}}{2}.\sqrt 2 = \dfrac{{ - 11\sqrt 2 }}{2}\\
b,\\
\sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 2} \right)}^2}} \\
= \left| {1 - \sqrt 3 } \right| - \left| {\sqrt 3 + 2} \right|\\
= \left( {\sqrt 3 - 1} \right) - \left( {\sqrt 3 + 2} \right)\\
= \sqrt 3 - 1 - \sqrt 3 - 2\\
= - 3\\
c,\\
\sqrt {48} + \sqrt {5\dfrac{1}{3}} + 2\sqrt {75} - 5\sqrt {1\dfrac{1}{3}} \\
= \sqrt {16.3} + \sqrt {\dfrac{{16}}{3}} + 2\sqrt {25.3} - 5\sqrt {\dfrac{4}{3}} \\
= \sqrt {{4^2}.3} + \dfrac{{\sqrt {16} }}{{\sqrt 3 }} + 2.\sqrt {{5^2}.3} - 5.\dfrac{{\sqrt 4 }}{{\sqrt 3 }}\\
= 4\sqrt 3 + \dfrac{4}{{\sqrt 3 }} + 2.5\sqrt 3 - 5.\dfrac{2}{{\sqrt 3 }}\\
= 4\sqrt 3 + \dfrac{{4\sqrt 3 }}{3} + 10\sqrt 3 - \dfrac{{10\sqrt 3 }}{3}\\
= \sqrt 3 .\left( {4 + \dfrac{4}{3} + 10 - \dfrac{{10}}{3}} \right)\\
= 12\sqrt 3 \\
d,\\
\dfrac{2}{{1 - \sqrt 2 }} - \dfrac{2}{{1 + \sqrt 2 }}\\
= \dfrac{{2.\left( {1 + \sqrt 2 } \right) - 2.\left( {1 - \sqrt 2 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}\\
= \dfrac{{2 + 2\sqrt 2 - 2 + 2\sqrt 2 }}{{1 - 2}}\\
= \dfrac{{4\sqrt 2 }}{{ - 1}}\\
= - 4\sqrt 2
\end{array}\)
