Đáp án:
\[A = - \dfrac{7}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \left( {\dfrac{1}{{3 - 2\sqrt 2 }} + \dfrac{2}{{\sqrt 2 - 2}} - \dfrac{3}{{2\sqrt 2 }}} \right).\left( {\sqrt 2 - 4} \right)\\
= \left( {\dfrac{{3 + 2\sqrt 2 }}{{\left( {3 - 2\sqrt 2 } \right)\left( {3 + 2\sqrt 2 } \right)}} + \dfrac{{2.\left( {\sqrt 2 + 2} \right)}}{{\left( {\sqrt 2 - 2} \right)\left( {\sqrt 2 + 2} \right)}} - \dfrac{{3.\sqrt 2 }}{{2\sqrt 2 .\sqrt 2 }}} \right).\left( {\sqrt 2 - 4} \right)\\
= \left( {\dfrac{{3 + 2\sqrt 2 }}{{{3^2} - {{\left( {2\sqrt 2 } \right)}^2}}} + \dfrac{{2\sqrt 2 + 4}}{{{{\sqrt 2 }^2} - {2^2}}} - \dfrac{{3\sqrt 2 }}{{2.2}}} \right).\left( {\sqrt 2 - 4} \right)\\
= \left( {\dfrac{{3 + 2\sqrt 2 }}{{9 - 8}} + \dfrac{{2\sqrt 2 + 4}}{{2 - 4}} - \dfrac{{3\sqrt 2 }}{4}} \right).\left( {\sqrt 2 - 4} \right)\\
= \left( {\dfrac{{3 + 2\sqrt 2 }}{1} + \dfrac{{2\sqrt 2 + 4}}{{ - 2}} - \dfrac{3}{4}\sqrt 2 } \right).\left( {\sqrt 2 - 4} \right)\\
= \left( {3 + 2\sqrt 2 - \sqrt 2 - 2 - \dfrac{3}{4}\sqrt 2 } \right).\left( {\sqrt 2 - 4} \right)\\
= \left( {1 + \dfrac{1}{4}\sqrt 2 } \right)\left( {\sqrt 2 - 4} \right)\\
= \dfrac{1}{4}.\left( {4 + \sqrt 2 } \right).\left( {\sqrt 2 - 4} \right)\\
= \dfrac{1}{4}.\left( {{{\sqrt 2 }^2} - {4^2}} \right)\\
= \dfrac{1}{4}.\left( {2 - 16} \right)\\
= \dfrac{1}{4}.\left( { - 14} \right)\\
= \dfrac{{ - 7}}{2}
\end{array}\)
