Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\sqrt {2005} \\
b,\,\,\,\,16
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {\sqrt {27} - 4\sqrt 3 + \sqrt {12} } \right).\sqrt 3 + \sqrt {{{\left( {3 - \sqrt {2005} } \right)}^2}} \\
= \left( {\sqrt {{3^2}.3} - 4\sqrt 3 + \sqrt {{2^2}.3} } \right).\sqrt 3 + \left| {3 - \sqrt {2005} } \right|\\
= \left( {3\sqrt 3 - 4\sqrt 3 + 2\sqrt 3 } \right).\sqrt 3 + \left( {\sqrt {2005} - 3} \right)\\
= \sqrt 3 .\sqrt 3 + \sqrt {2005} - 3\\
= 3 + \sqrt {2005} - 3\\
= \sqrt {2005} \\
b,\\
\dfrac{{12 - 4\sqrt 3 }}{{\sqrt 3 - 1}} + 12\sqrt {\dfrac{1}{3}} + \dfrac{8}{{\sqrt 3 + 2}}\\
= \dfrac{{4.\left( {3 - \sqrt 3 } \right)}}{{\sqrt 3 - 1}} + 12.\sqrt {\dfrac{1}{9}.3} + \dfrac{{8\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}\\
= \dfrac{{4.\left( {{{\sqrt 3 }^2} - \sqrt 3 } \right)}}{{\sqrt 3 - 1}} + 12.\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}.3} + \dfrac{{8.\left( {2 - \sqrt 3 } \right)}}{{{2^2} - {{\sqrt 3 }^2}}}\\
= \dfrac{{4.\sqrt 3 .\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}} + 12.\dfrac{1}{3}\sqrt 3 + \dfrac{{8.\left( {2 - \sqrt 3 } \right)}}{1}\\
= 4\sqrt 3 + 4\sqrt 3 + 8.\left( {2 - \sqrt 3 } \right)\\
= 8\sqrt 3 + 16 - 8\sqrt 3 \\
= 16
\end{array}\)
