A=( $\frac{1}{2+\sqrt3}$ – $\frac{12}{3+ \sqrt 3}$ +$\frac{26}{4- \sqrt3}$ ) ($\sqrt{4}$ -3 $\sqrt{3}$)

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1. Đáp án:

   $\begin{array}{l}
   A = \left( {\dfrac{1}{{2 + \sqrt 3 }} – \dfrac{{12}}{{3 + \sqrt 3 }} + \dfrac{{26}}{{4 – \sqrt 3 }}} \right)\left( {\sqrt 4  – 3\sqrt 3 } \right)\\
    = \left( \begin{array}{l}
   \dfrac{{2 – \sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 – \sqrt 3 } \right)}} – \dfrac{{12\left( {3 – \sqrt 3 } \right)}}{{\left( {3 + \sqrt 3 } \right)\left( {3 – \sqrt 3 } \right)}}\\
    + \dfrac{{26\left( {4 + \sqrt 3 } \right)}}{{\left( {4 + \sqrt 3 } \right)\left( {4 – \sqrt 3 } \right)}}
   \end{array} \right)\\
   .\left( {2 – 3\sqrt 3 } \right)\\
    = \left( {\dfrac{{2 – \sqrt 3 }}{{4 – 3}} – \dfrac{{36 – 12\sqrt 3 }}{{9 – 3}} + \dfrac{{104 + 26\sqrt 3 }}{{16 – 3}}} \right).\left( {2 – 3\sqrt 3 } \right)\\
    = \left( {2 – \sqrt 3  – 6 + 2\sqrt 3  + 8 + 2\sqrt 3 } \right).\left( {2 – 3\sqrt 3 } \right)\\
    = \left( {4 + 3\sqrt 3 } \right)\left( {2 – 3\sqrt 3 } \right)\\
    = 8 – 12\sqrt 3  + 6\sqrt 3  – 27\\
    =  – 19 – 6\sqrt 3 
   \end{array}$

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