Đáp án:
\(\begin{array}{l}
a,\\
A = \sqrt 3 - 5\\
b,\\
B = - 2\\
c,\\
C = 1 - 4\sqrt 3 \\
d,\\
D = 3\sqrt 3 - \sqrt 5 + 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
A = \dfrac{6}{{\sqrt 3 + 1}} - \dfrac{4}{{\sqrt 3 - 1}}\\
= \dfrac{{6.\left( {\sqrt 3 - 1} \right) - 4.\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{{6\sqrt 3 - 6 - 4\sqrt 3 - 4}}{{{{\sqrt 3 }^2} - {1^2}}}\\
= \dfrac{{2\sqrt 3 - 10}}{{3 - 1}}\\
= \dfrac{{2\sqrt 3 - 10}}{2}\\
= \sqrt 3 - 5\\
b,\\
B = \dfrac{1}{{1 - \sqrt 2 }} + \dfrac{1}{{1 + \sqrt 2 }}\\
= \dfrac{{1.\left( {1 + \sqrt 2 } \right) + 1.\left( {1 - \sqrt 2 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}\\
= \dfrac{{1 + \sqrt 2 + 1 - \sqrt 2 }}{{{1^2} - {{\sqrt 2 }^2}}}\\
= \dfrac{2}{{1 - 2}}\\
= \dfrac{2}{{ - 1}}\\
= - 2\\
c,\\
C = \dfrac{2}{{\sqrt 3 + 2}} - \dfrac{3}{{\sqrt 2 - 1}} - \left( {2\sqrt 3 - 3\sqrt 2 } \right)\\
= \dfrac{{2.\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} - \dfrac{{3.\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} - 2\sqrt 3 + 3\sqrt 2 \\
= \dfrac{{4 - 2\sqrt 3 }}{{{2^2} - {{\sqrt 3 }^2}}} - \dfrac{{3\sqrt 2 + 3}}{{{{\sqrt 2 }^2} - {1^2}}} - 2\sqrt 3 + 3\sqrt 2 \\
= \dfrac{{4 - 2\sqrt 3 }}{{4 - 3}} - \dfrac{{3\sqrt 2 + 3}}{{2 - 1}} - 2\sqrt 3 + 3\sqrt 2 \\
= \dfrac{{4 - 2\sqrt 3 }}{1} - \dfrac{{3\sqrt 2 + 3}}{1} - 2\sqrt 3 + 3\sqrt 2 \\
= 4 - 2\sqrt 3 - 3\sqrt 2 - 3 - 2\sqrt 3 + 3\sqrt 2 \\
= 1 - 4\sqrt 3 \\
d,\\
D = \dfrac{{26}}{{3\sqrt 3 - 1}} - \dfrac{4}{{\sqrt 5 + 1}}\\
= \dfrac{{26\left( {3\sqrt 3 + 1} \right)}}{{\left( {3\sqrt 3 - 1} \right)\left( {3\sqrt 3 + 1} \right)}} - \dfrac{{4.\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{26.\left( {3\sqrt 3 + 1} \right)}}{{{{\left( {3\sqrt 3 } \right)}^2} - {1^2}}} - \dfrac{{4.\left( {\sqrt 5 - 1} \right)}}{{{{\sqrt 5 }^2} - {1^2}}}\\
= \dfrac{{26.\left( {3\sqrt 3 + 1} \right)}}{{27 - 1}} - \dfrac{{4.\left( {\sqrt 5 - 1} \right)}}{{5 - 1}}\\
= \dfrac{{26.\left( {3\sqrt 3 + 1} \right)}}{{26}} - \dfrac{{4\left( {\sqrt 5 - 1} \right)}}{4}\\
= \left( {3\sqrt 3 + 1} \right) - \left( {\sqrt 5 - 1} \right)\\
= 3\sqrt 3 + 1 - \sqrt 5 + 1\\
= 3\sqrt 3 - \sqrt 5 + 2
\end{array}\)
