Bài 2 trang 36 Tài liệu dạy – học Toán 9 tập 1>
Giải bài tập Tính :
Đề bài
Tính :
a) \(\sqrt[3]{{64}} - \sqrt[3]{{ - 125}} - 2\sqrt[3]{{216}}\);
b) \(\sqrt[3]{{81}} + 3\sqrt[3]{{24}} - 2\sqrt[3]{{375}}\);
c) \(\dfrac{{\sqrt[3]{{135}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{54}}.\sqrt[3]{4}\);
d) \(\sqrt[3]{{54x{y^3}}} - y\sqrt[3]{{128x}}\);
e) \(\sqrt {7 - 4\sqrt 3 } - \sqrt[3]{{26 + 15\sqrt 3 }}\).
Phương pháp giải - Xem chi tiết
+) Sử dụng công thức: \(\sqrt[3]{{ab}} = \sqrt[3]{a}.\sqrt[3]{b},\;\;\sqrt[3]{{\dfrac{a}{b}}} = \dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}}\;\;\left( {b \ne 0} \right).\)
Lời giải chi tiết
\(\begin{array}{l}a)\;\sqrt[3]{{64}} - \sqrt[3]{{ - 125}} - 2\sqrt[3]{{216}}\\ = \sqrt[3]{{{4^3}}} - \sqrt[3]{{{{\left( { - 5} \right)}^3}}} - 2\sqrt[3]{{{6^3}}}\\ = 4 - \left( { - 5} \right) - 2.6\\ = - 3.\end{array}\) \(\begin{array}{l}b)\;\sqrt[3]{{81}} + 3\sqrt[3]{{24}} - 2\sqrt[3]{{375}}\\ = \sqrt[3]{{{{3.3}^3}}} + 3\sqrt[3]{{{{3.2}^3}}} - 2\sqrt[3]{{{{3.5}^3}}}\\ = 3\sqrt[3]{3} + 3.2\sqrt[3]{3} - 2.5\sqrt[3]{3}\\ = - \sqrt[3]{3}.\end{array}\)
\(\begin{array}{l}c)\;\dfrac{{\sqrt[3]{{135}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{54}}.\sqrt[3]{4}\\ = \sqrt[3]{{\dfrac{{135}}{5}}} - \sqrt[3]{{54.4}}\\ = \sqrt[3]{{27}} - \sqrt[3]{{216}}\\ = \sqrt[3]{{{3^3}}} - \sqrt[3]{{{6^3}}}\\ = 3 - 6 = - 3.\end{array}\) \(\begin{array}{l}d)\;\sqrt[3]{{54x{y^3}}} - y\sqrt[3]{{128x}}\\ = \sqrt[3]{{{{2.3}^3}x{y^3}}} - y\sqrt[3]{{{{2.4}^3}x}}\\ = 3y\sqrt[3]{{2x}} - 4y\sqrt[3]{{2x}}\\ = - y\sqrt[3]{{2x}}\end{array}\)
\(e)\;\sqrt {7 - 4\sqrt 3 } - \sqrt[3]{{26 + 15\sqrt 3 }}\)
\(= \sqrt {4 - 2.2.\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} \)\(\; - \sqrt[3]{{{2^3} + {{3.2}^2}\sqrt 3 + 3.2.{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^3}}}\)
\( = \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} - \sqrt[3]{{{{\left( {2 + \sqrt 3 } \right)}^3}}}\\ = \left| {2 - \sqrt 3 } \right| - \left( {2 + \sqrt 3 } \right)\)
\( = 2 - \sqrt 3 - 2 - \sqrt 3 = - 2\sqrt 3 .\)
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