Đáp án:
\(\begin{array}{l}
B1:\\
2\sqrt 6 \\
B2:\\
A = \dfrac{{2\sqrt x }}{{\sqrt x - 2}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
d)\dfrac{{\left( {3 + \sqrt 3 } \right)\sqrt {4 - 2\sqrt 3 } + \left( {3 - \sqrt 3 } \right)\sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 }}\\
= \dfrac{{\left( {3 + \sqrt 3 } \right)\sqrt {3 - 2\sqrt 3 .1 + 1} + \left( {3 - \sqrt 3 } \right)\sqrt {3 + 2\sqrt 3 .1 + 1} }}{{\sqrt 2 }}\\
= \dfrac{{\left( {3 + \sqrt 3 } \right)\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \left( {3 - \sqrt 3 } \right)\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\left( {3 + \sqrt 3 } \right)\left( {\sqrt 3 - 1} \right) + \left( {3 - \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)}}{{\sqrt 2 }}\\
= \dfrac{{3\sqrt 3 - 3 + 3 - \sqrt 3 + 3\sqrt 3 + 3 - 3 - \sqrt 3 }}{{\sqrt 2 }}\\
= \dfrac{{4\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{4\sqrt 6 }}{2} = 2\sqrt 6 \\
B2:\\
A = \left[ {\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}} + \dfrac{8}{{\sqrt x - 2}}} \right]:\dfrac{{4 + x + 4\sqrt x }}{{2x}}\\
= \left( {\dfrac{{\sqrt x - 2}}{{\sqrt x }} + \dfrac{8}{{\sqrt x - 2}}} \right).\dfrac{{2x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \dfrac{{{{\left( {\sqrt x - 2} \right)}^2} + 8\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{2x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \dfrac{{x - 4\sqrt x + 4 + 8\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{2x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \dfrac{{x + 4\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{2x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{2x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x - 2}}
\end{array}\)
