Đáp án:
\(\begin{array}{l}
a,\\
A = - 3\sqrt 2 - 9\sqrt 3 \\
b,\\
B = \sqrt 6 - \sqrt 5 \\
c,\\
C = - 2 - 2\sqrt 2 \\
d,\\
D = 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 3\sqrt {50} - 2\sqrt {75} - 3\sqrt {72} + 4\sqrt {48} - 5\sqrt {27} \\
= 3\sqrt {25.2} - 2.\sqrt {25.3} - 3\sqrt {36.2} + 4\sqrt {16.3} - 5\sqrt {9.3} \\
= 3\sqrt {{5^2}.2} - 2\sqrt {{5^2}.3} - 3\sqrt {{6^2}.2} + 4\sqrt {{4^2}.3} - 5\sqrt {{3^2}.3} \\
= 3.5\sqrt 2 - 2.5\sqrt 3 - 3.6\sqrt 2 + 4.4\sqrt 3 - 5.3\sqrt 3 \\
= 15\sqrt 2 - 10\sqrt 3 - 18\sqrt 2 + 16\sqrt 3 - 15\sqrt 3 \\
= \left( {15\sqrt 2 - 18\sqrt 2 } \right) + \left( { - 10\sqrt 3 + 16\sqrt 3 - 15\sqrt 3 } \right)\\
= - 3\sqrt 2 - 9\sqrt 3 \\
b,\\
B = \dfrac{2}{{\sqrt 5 - \sqrt 3 }} - \sqrt {20} + \dfrac{3}{{\sqrt 3 + \sqrt 6 }}\\
= \dfrac{{2\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}} - \sqrt {4.5} + \dfrac{{3.\left( {\sqrt 6 - \sqrt 3 } \right)}}{{\left( {\sqrt 6 + \sqrt 3 } \right)\left( {\sqrt 6 - \sqrt 3 } \right)}}\\
= \dfrac{{2.\left( {\sqrt 5 + \sqrt 3 } \right)}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} - \sqrt {{2^2}.5} + \dfrac{{3.\left( {\sqrt 6 - \sqrt 3 } \right)}}{{{{\sqrt 6 }^2} - {{\sqrt 3 }^2}}}\\
= \dfrac{{2.\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} - 2\sqrt 5 + \dfrac{{3.\left( {\sqrt 6 - \sqrt 3 } \right)}}{{6 - 3}}\\
= \dfrac{{2.\left( {\sqrt 5 + \sqrt 3 } \right)}}{2} - 2\sqrt 5 + \dfrac{{3.\left( {\sqrt 6 - \sqrt 3 } \right)}}{3}\\
= \left( {\sqrt 5 + \sqrt 3 } \right) - 2\sqrt 5 + \left( {\sqrt 6 - \sqrt 3 } \right)\\
= \sqrt 5 + \sqrt 3 - 2\sqrt 5 + \sqrt 6 - \sqrt 3 \\
= \sqrt 6 - \sqrt 5 \\
c,\\
C = \sqrt {3 + 2\sqrt 2 } - \sqrt {17 - 12\sqrt 2 } - 5\sqrt 2 \\
= \sqrt {2 + 2\sqrt 2 + 1} - \sqrt {9 - 12\sqrt 2 + 8} - 5\sqrt 2 \\
= \sqrt {{{\sqrt 2 }^2} + 2.\sqrt 2 .1 + {1^2}} - \sqrt {{3^2} - 2.3.2\sqrt 2 + {{\left( {2\sqrt 2 } \right)}^2}} - 5\sqrt 2 \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} - 5\sqrt 2 \\
= \left| {\sqrt 2 + 1} \right| - \left| {3 - 2\sqrt 2 } \right| - 5\sqrt 2 \\
= \left( {\sqrt 2 + 1} \right) - \left( {3 - 2\sqrt 2 } \right) - 5\sqrt 2 \\
= \sqrt 2 + 1 - 3 + 2\sqrt 2 - 5\sqrt 2 \\
= - 2 - 2\sqrt 2 \\
d,\\
D = \left( {\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{2 + \sqrt 2 }}{{\sqrt 2 + 1}}} \right) - \left( {1:\dfrac{1}{{\sqrt 2 + \sqrt 3 }}} \right)\\
= \left( {\dfrac{{{{\sqrt 3 }^2} + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{{{\sqrt 2 }^2} + \sqrt 2 }}{{\sqrt 2 + 1}}} \right) - 1.\left( {\sqrt 2 + \sqrt 3 } \right)\\
= \left( {\dfrac{{\sqrt 3 .\left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} + \dfrac{{\sqrt 2 .\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}}} \right) - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \left[ {\left( {\sqrt 3 + 2} \right) + \sqrt 2 } \right] - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \sqrt 3 + 2 + \sqrt 2 - \sqrt 2 - \sqrt 3 \\
= 2
\end{array}\)
