Bài 3 trang 39 Tài liệu dạy – học Toán 9 tập 1


    Đề bàiThực hiện phép tính:a) \(\sqrt {50}  - 3\sqrt {98}  + 2\sqrt 8  + \sqrt {32}  - 5\sqrt {18} \);b) \(\dfrac{1}{2}\sqrt {48}  - 2\sqrt {75}  - \sqrt {108}  + 5\sqrt {1\dfrac{1}{3}} \);c) \(\dfrac{{2\sqrt 3  - 3\sqrt 2 }}{{\sqrt 3  - \sqrt 2 }} + \dfrac{5}{{1 + \sqrt 6 }}\); d) \(\dfrac{3}{{\sqrt...

    Đề bài

    Thực hiện phép tính:

    a) \(\sqrt {50}  – 3\sqrt {98}  + 2\sqrt 8  + \sqrt {32}  – 5\sqrt {18} \);

    b) \(\dfrac{1}{2}\sqrt {48}  – 2\sqrt {75}  – \sqrt {108}  + 5\sqrt {1\dfrac{1}{3}} \);

    c) \(\dfrac{{2\sqrt 3  – 3\sqrt 2 }}{{\sqrt 3  – \sqrt 2 }} + \dfrac{5}{{1 + \sqrt 6 }}\); 

    d) \(\dfrac{3}{{\sqrt 7  + 2}} + \sqrt {\dfrac{2}{{8 + 3\sqrt 7 }}} \);                

    e) \(\dfrac{1}{{\sqrt 2  – \sqrt 3 }}\sqrt {\dfrac{{3\sqrt 2  – 2\sqrt 3 }}{{3\sqrt 2  + 2\sqrt 3 }}} \);   

    f) \(\dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {13 + \sqrt {48} } } } }}{{\sqrt 6  + \sqrt 2 }}\).

    Phương pháp giải – Xem chi tiếtBài 3 trang 39 Tài liệu dạy – học Toán 9 tập 1

    +) 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

    \(a)\;\sqrt {50}  – 3\sqrt {98}  + 2\sqrt 8  + \sqrt {32}  – 5\sqrt {18} \)

    \(= \sqrt {{5^2}.2}  – 3\sqrt {{7^2}.2}  + 2\sqrt {{2^2}.2}  + \sqrt {{4^2}.2}\)\(\,  – 5\sqrt {{3^2}.2} \)

    \(= 5\sqrt 2  – 21\sqrt 2  + 4\sqrt 2  + 4\sqrt 2  – 15\sqrt 2 \)

    \(=  – 23\sqrt 2 \)

    \(\begin{array}{l}b)\;\dfrac{1}{2}\sqrt {48}  – 2\sqrt {75}  – \sqrt {108}  + 5\sqrt {1\dfrac{1}{3}} \\ = \dfrac{1}{2}\sqrt {{4^2}.3}  – 2\sqrt {{5^2}.3}  – \sqrt {{6^2}.3}  + 5\sqrt {\dfrac{4}{3}} \\ = \dfrac{1}{2}.2\sqrt 3  – 2.5\sqrt 3  – 6\sqrt 3  + 5.\sqrt {\dfrac{{4.3}}{{{3^2}}}} \\ =  – 15\sqrt 3  + 5.\dfrac{4}{3}\sqrt 3  =  – 15\sqrt 3  + \dfrac{{20\sqrt 3 }}{3}\\ =  – \dfrac{{25\sqrt 3 }}{3}.\end{array}\)

    \(\begin{array}{l}c)\;\dfrac{{2\sqrt 3  – 3\sqrt 2 }}{{\sqrt 3  – \sqrt 2 }} + \dfrac{5}{{1 + \sqrt 6 }}\\ = \dfrac{{\sqrt 6 \left( {\sqrt 2  – \sqrt 3 } \right)}}{{\sqrt 3  – \sqrt 2 }} + \dfrac{{5\left( {1 – \sqrt 6 } \right)}}{{1 – 6}}\\ =  – \sqrt 6  + \dfrac{{5\left( {1 – \sqrt 6 } \right)}}{{ – 5}}\\ =  – \sqrt 6  – \left( {1 – \sqrt 6 } \right)\\ =  – \sqrt 6  + \sqrt 6  – 1 =  – 1.\end{array}\)

    \(\begin{array}{l}d)\;\dfrac{3}{{\sqrt 7  + 2}} + \sqrt {\dfrac{2}{{8 + 3\sqrt 7 }}} \\ = \dfrac{{3\left( {\sqrt 7  – 2} \right)}}{{7 – 4}} + \sqrt {\dfrac{{2\left( {8 – 3\sqrt 7 } \right)}}{{{8^2} – {{\left( {3\sqrt 7 } \right)}^2}}}} \\ = \dfrac{{3\left( {\sqrt 7  – 2} \right)}}{3} + \sqrt {\dfrac{{16 – 6\sqrt 7 }}{1}} \\ = \sqrt 7  – 2 + \sqrt {{3^2} – 2.3\sqrt 7  + {{\left( {\sqrt 7 } \right)}^2}} \\ = \sqrt 7  – 2 + \sqrt {{{\left( {3 – \sqrt 7 } \right)}^2}} \\ = \sqrt 7  – 2 + \left| {3 – \sqrt 7 } \right|\\ = \sqrt 7  – 2 + 3 – \sqrt 7 \\ = 3 – 2 = 1.\end{array}\)

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

    \(\begin{array}{l}f)\;\dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {13 + \sqrt {48} } } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {13 + \sqrt {{4^2}.3} } } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {13 + 4\sqrt 3 } } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {13 + 4\sqrt 3 } } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + 2.2\sqrt 3  + 1} } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – \sqrt {{{\left( {2\sqrt 3  + 1} \right)}^2}} } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – \left| {2\sqrt 3  + 1} \right|} } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {5 – 2\sqrt 3  – 1} } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {4 – 2\sqrt 3 } } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {{{\left( {\sqrt 3 } \right)}^2} – 2\sqrt 3  + 1} } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \sqrt {{{\left( {\sqrt 3  – 1} \right)}^2}} } }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {3 + \left| {\sqrt 3  – 1} \right|} }}{{\sqrt 6  + \sqrt 2 }} = \dfrac{{2\sqrt {3 + \sqrt 3  – 1} }}{{\sqrt 6  + \sqrt 2 }}\\ = \dfrac{{2\sqrt {2 + \sqrt 3 } }}{{\sqrt 6  + \sqrt 2 }} = \dfrac{{\sqrt 2 \sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 \left( {\sqrt 3  + 1} \right)}}\\ = \dfrac{{\sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2\sqrt 3  + 1} }}{{\sqrt 3  + 1}} = \dfrac{{\sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}} }}{{\sqrt 3  + 1}}\\ = \dfrac{{\left| {\sqrt 3  + 1} \right|}}{{\sqrt 3  + 1}} = \dfrac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}} = 1.\end{array}\)