Bài 11 trang 30 Tài liệu dạy – học Toán 9 tập 1

Giải bài tập Rút gọn :

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

\(\begin{array}{l}a)\;\dfrac{1}{{\sqrt {2 - \sqrt 3 } }} + \dfrac{1}{{\sqrt {2 + \sqrt 3 } }}\\ = \dfrac{{\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } }}{{\sqrt {2 - \sqrt 3 } .\sqrt {2 + \sqrt 3 } }}\\ = \dfrac{{\sqrt 2 \left( {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right)}}{{\sqrt 2 \left( {\sqrt {4 - 3} } \right)}}\\ = \dfrac{{\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{{\sqrt 2 }}\\ = \dfrac{{\left| {\sqrt 3 + 1} \right| + \left| {\sqrt 3 - 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{{\sqrt 2 }}\\ = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 3 = \sqrt 6 .\end{array}\)

\(\begin{array}{l}b)\;\sqrt {32 - 4\sqrt {60} } - \sqrt {\dfrac{2}{{4 + \sqrt {15} }}} \\ = \sqrt {32 - 4\sqrt {{2^2}.15} } - \sqrt {\dfrac{{2\left( {4 - \sqrt {15} } \right)}}{{{4^2} - 15}}} \\ = \sqrt {32 - 8\sqrt {15} } - \sqrt {8 - 2\sqrt {15} } \\ = \sqrt {{{\left( {2\sqrt 5 } \right)}^2} - 2.2\sqrt 5 .2\sqrt 3 + {{\left( {2\sqrt 3 } \right)}^2}} \\\;\;\;\; - \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 5 .\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} \\ = \sqrt {{{\left( {2\sqrt 5 - 2\sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\ = \left| {2\sqrt 5 - 2\sqrt 3 } \right| - \left| {\sqrt 5 - \sqrt 3 } \right|\\ = 2\sqrt 5 - 2\sqrt 3 - \sqrt 5 + \sqrt 3 \\ = \sqrt 5 - \sqrt 3 .\end{array}\)

\(\begin{array}{l}c)\;\dfrac{{\sqrt 3 }}{{3 - \sqrt 3 }} + \dfrac{{\sqrt 6 }}{{\sqrt {6 + 3\sqrt 3 } }}\\ = \dfrac{{\sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}} + \dfrac{{\sqrt 6 }}{{\sqrt {3\left( {2 + \sqrt 3 } \right)} }}\\ = \dfrac{1}{{\sqrt 3 - 1}} + \dfrac{{\sqrt 2 }}{{\sqrt {2 + \sqrt 3 } }}\\ = \dfrac{{\sqrt 3 + 1}}{{3 - 1}} + \dfrac{{\sqrt 2 .\sqrt 2 }}{{\sqrt {2\left( {2 + \sqrt 3 } \right)} }}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{2}{{\sqrt {4 + 2\sqrt 3 } }}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{2}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{2}{{\sqrt 3 + 1}}\\ = \dfrac{{\sqrt 3 + 1}}{2} + \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}}\\ = \dfrac{{\sqrt 3 + 1 + 2\sqrt 3 - 2}}{2}\\ = \dfrac{{3\sqrt 3 - 1}}{2}.\end{array}\)

\(\begin{array}{l}d)\;\sqrt {\dfrac{{3\sqrt 3 - 4}}{{2\sqrt 3 + 1}}} + \sqrt {\dfrac{{\sqrt 3 + 4}}{{5 - 2\sqrt 3 }}} \\ = \sqrt {\dfrac{{\left( {3\sqrt 3 - 4} \right)\left( {2\sqrt 3 - 1} \right)}}{{{{\left( {2\sqrt 3 } \right)}^2} - 1}}} \\\;\;\;\; + \sqrt {\dfrac{{\left( {\sqrt 3 + 4} \right)\left( {5 + 2\sqrt 3 } \right)}}{{{5^2} - {{\left( {2\sqrt 3 } \right)}^2}}}} \\ = \sqrt {\dfrac{{22 - 11\sqrt 3 }}{{11}}} + \sqrt {\dfrac{{26 + 13\sqrt 3 }}{{13}}} \\ = \sqrt {2 - \sqrt 3 } + \sqrt {2 + \sqrt 3 } \\ = \dfrac{{\sqrt 2 \left( {\sqrt {2 - \sqrt 3 } + \sqrt {2 + \sqrt 3 } } \right)}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}{{\sqrt 2 }}\\ = \dfrac{{\left| {\sqrt 3 - 1} \right| + \left| {\sqrt 3 + 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 3 - 1 + \sqrt 3 + 1}}{{\sqrt 2 }}\\ = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 3 = \sqrt 6 .\end{array}\)

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