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

Giải bài tập Thực hiện phép tính :

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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 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

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

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

\(= 5\sqrt 2 - 3.7\sqrt 2 + 2.2\sqrt 2 + 4\sqrt 2 - 5.3\sqrt 2 \)

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

\(=\sqrt 2 .(5-21+4+4-15)\)

\(= - 23\sqrt 2 \)

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

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

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

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

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

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