Đề bài
Rút gọn :
a] \[\dfrac{1}{{\sqrt {2 - \sqrt 3 } }} + \dfrac{1}{{\sqrt {2 + \sqrt 3 } }}\];
b] \[\sqrt {32 - 4\sqrt {60} } - \sqrt {\dfrac{2}{{4 + \sqrt {15} }}} \];
c] \[\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} + \dfrac{{\sqrt 6 }}{{\sqrt {6 + 3\sqrt 3 } }}\];
d] \[\sqrt {\dfrac{{3\sqrt 3 - 4}}{{2\sqrt 3 + 1}}} + \sqrt {\dfrac{{\sqrt 3 + 4}}{{5 - 2\sqrt 3 }}} \].
Phương pháp giải - Xem chi tiết
+] Sử dụng công thức trục căn thức ở mẫu:\[\sqrt {\dfrac{A}{B}} = \sqrt {\dfrac{{A.B}}{{{B^2}}}} = \dfrac{{\sqrt {AB} }}{B},\]\[\;\;A\sqrt {\dfrac{B}{A}} = \sqrt {\dfrac{{{A^2}.B}}{A}} = \sqrt {AB} .\]
+] \[\dfrac{C}{{\sqrt A \pm B}} = \dfrac{{C\left[ {\sqrt A \mp B} \right]}}{{A - {B^2}}};\]\[\;\;\dfrac{C}{{\sqrt A \pm \sqrt B }} = \dfrac{{C\left[ {\sqrt A \mp \sqrt B } \right]}}{{A - B}}.\]
+] \[\sqrt {{A^2}B} = \left| A \right|\sqrt B = \left\{ \begin{array}{l}A\sqrt B \;\;khi\;\;A \ge 0\\ - A\sqrt B \;\;khi\;\;A < 0\end{array} \right..\]
Lời giải chi tiết
\[\begin{array}{l}a]\;\dfrac{1}{{\sqrt {2 - \sqrt 3 } }} + \dfrac{1}{{\sqrt {2 + \sqrt 3 } }}\\ = \dfrac{{\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } }}{{\sqrt {2 - \sqrt 3 } .\sqrt {2 + \sqrt 3 } }}\\ = \dfrac{{\sqrt 2 \left[ {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right]}}{{\sqrt 2 \left[ {\sqrt {4 - 3} } \right]}}\\ = \dfrac{{\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left[ {\sqrt 3 + 1} \right]}^2}} + \sqrt {{{\left[ {\sqrt 3 - 1} \right]}^2}} }}{{\sqrt 2 }}\\ = \dfrac{{\left| {\sqrt 3 + 1} \right| + \left| {\sqrt 3 - 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{{\sqrt 2 }}\\ = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 3 = \sqrt 6 .\end{array}\]
\[\begin{array}{l}b]\;\sqrt {32 - 4\sqrt {60} } - \sqrt {\dfrac{2}{{4 + \sqrt {15} }}} \\ = \sqrt {32 - 4\sqrt {{2^2}.15} } - \sqrt {\dfrac{{2\left[ {4 - \sqrt {15} } \right]}}{{{4^2} - 15}}} \\ = \sqrt {32 - 8\sqrt {15} } - \sqrt {8 - 2\sqrt {15} } \\ = \sqrt {{{\left[ {2\sqrt 5 } \right]}^2} - 2.2\sqrt 5 .2\sqrt 3 + {{\left[ {2\sqrt 3 } \right]}^2}} \\\;\;\;\; - \sqrt {{{\left[ {\sqrt 5 } \right]}^2} - 2.\sqrt 5 .\sqrt 3 + {{\left[ {\sqrt 3 } \right]}^2}} \\ = \sqrt {{{\left[ {2\sqrt 5 - 2\sqrt 3 } \right]}^2}} - \sqrt {{{\left[ {\sqrt 5 - \sqrt 3 } \right]}^2}} \\ = \left| {2\sqrt 5 - 2\sqrt 3 } \right| - \left| {\sqrt 5 - \sqrt 3 } \right|\\ = 2\sqrt 5 - 2\sqrt 3 - \sqrt 5 + \sqrt 3 \\ = \sqrt 5 - \sqrt 3 .\end{array}\]
\[\begin{array}{l}c]\;\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} + \dfrac{{\sqrt 6 }}{{\sqrt {6 + 3\sqrt 3 } }}\\ = \dfrac{{\sqrt 3 }}{{\sqrt 3 \left[ {\sqrt 3 - 1} \right]}} + \dfrac{{\sqrt 6 }}{{\sqrt {3\left[ {2 + \sqrt 3 } \right]} }}\\ = \dfrac{1}{{\sqrt 3 - 1}} + \dfrac{{\sqrt 2 }}{{\sqrt {2 + \sqrt 3 } }}\\ = \dfrac{{\sqrt 3 + 1}}{{3 - 1}} + \dfrac{{\sqrt 2 .\sqrt 2 }}{{\sqrt {2\left[ {2 + \sqrt 3 } \right]} }}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{2}{{\sqrt {4 + 2\sqrt 3 } }}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{2}{{\sqrt {{{\left[ {\sqrt 3 + 1} \right]}^2}} }}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{2}{{\sqrt 3 + 1}}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{{2\left[ {\sqrt 3 - 1} \right]}}{{3 - 1}}\\ = \dfrac{{\sqrt 3 + 1 + 2\sqrt 3 - 2}}{2}\\ = \dfrac{{3\sqrt 3 - 1}}{2}.\end{array}\]
\[\begin{array}{l}d]\;\sqrt {\dfrac{{3\sqrt 3 - 4}}{{2\sqrt 3 + 1}}} + \sqrt {\dfrac{{\sqrt 3 + 4}}{{5 - 2\sqrt 3 }}} \\ = \sqrt {\dfrac{{\left[ {3\sqrt 3 - 4} \right]\left[ {2\sqrt 3 - 1} \right]}}{{{{\left[ {2\sqrt 3 } \right]}^2} - 1}}} \\\;\;\;\; + \sqrt {\dfrac{{\left[ {\sqrt 3 + 4} \right]\left[ {5 + 2\sqrt 3 } \right]}}{{{5^2} - {{\left[ {2\sqrt 3 } \right]}^2}}}} \\ = \sqrt {\dfrac{{22 - 11\sqrt 3 }}{{11}}} + \sqrt {\dfrac{{26 + 13\sqrt 3 }}{{13}}} \\ = \sqrt {2 - \sqrt 3 } + \sqrt {2 + \sqrt 3 } \\ = \dfrac{{\sqrt 2 \left[ {\sqrt {2 - \sqrt 3 } + \sqrt {2 + \sqrt 3 } } \right]}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left[ {\sqrt 3 - 1} \right]}^2}} + \sqrt {{{\left[ {\sqrt 3 + 1} \right]}^2}} }}{{\sqrt 2 }}\\ = \dfrac{{\left| {\sqrt 3 - 1} \right| + \left| {\sqrt 3 + 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 3 - 1 + \sqrt 3 + 1}}{{\sqrt 2 }}\\ = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 3 = \sqrt 6 .\end{array}\]