Đáp án:
\(\begin{array}{l}
a,\\
A = 5\\
b,\\
B = 5\sqrt 3 + 1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{\sqrt 7 + \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }} + \dfrac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }}\\
= \dfrac{{{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2} + {{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}}}{{\left( {\sqrt 7 - \sqrt 3 } \right)\left( {\sqrt 7 + \sqrt 3 } \right)}}\\
= \dfrac{{\left( {{{\sqrt 7 }^2} + 2.\sqrt 7 .\sqrt 3 + {{\sqrt 3 }^2}} \right) + \left( {{{\sqrt 7 }^2} - 2.\sqrt 7 .\sqrt 3 + {{\sqrt 3 }^2}} \right)}}{{{{\sqrt 7 }^2} - {{\sqrt 3 }^2}}}\\
= \dfrac{{\left( {7 + 2\sqrt {21} + 3} \right) + \left( {7 - 2\sqrt {21} + 3} \right)}}{{7 - 3}}\\
= \dfrac{{10 + 2\sqrt {21} + 10 - 2\sqrt {21} }}{4}\\
= \dfrac{{20}}{4}\\
= 5\\
b,\\
B = 2\sqrt {27} + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} - \dfrac{4}{{\sqrt 3 + 1}}\\
= 2\sqrt {{3^2}.3} + \left| {1 - \sqrt 3 } \right| - \dfrac{{4\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\\
= 2.3\sqrt 3 + \left( {\sqrt 3 - 1} \right) - \dfrac{{4\left( {\sqrt 3 - 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}}\\
= 6\sqrt 3 + \sqrt 3 - 1 - \dfrac{{4.\left( {\sqrt 3 - 1} \right)}}{2}\\
= 7\sqrt 3 - 1 - 2.\left( {\sqrt 3 - 1} \right)\\
= 7\sqrt 3 - 1 - 2\sqrt 3 + 2\\
= 5\sqrt 3 + 1
\end{array}\)
