10 bài tập Toán 9 Kết nối tri thức Bài 10. Căn bậc ba và căn thức bậc ba có đáp án

a) \(\sqrt[3]{{64}} = 4\).

b) \(\sqrt[3]{{ - 512}} =  - 8\)

c) \(\sqrt[3]{{0,064}} = \sqrt[3]{{\frac{{64}}{{1000}}}} = \frac{{\sqrt[3]{{64}}}}{{\sqrt[3]{{1000}}}} = \frac{4}{{10}}\).

d) \(\sqrt[3]{{ - 0,216}} = \sqrt[3]{{{{\left( { - 6} \right)}^3}}} =  - 6\).

e) \(\frac{{\sqrt[3]{{500}}}}{{\sqrt[3]{4}}} + \sqrt[3]{{12}} \cdot \sqrt[3]{{18}} = \sqrt[3]{{\frac{{500}}{4}}} + \sqrt[3]{{12 \cdot 18}} = \sqrt[3]{{125}} + \sqrt[3]{{216}} = 5 + 6 = 11\)

f) \(\frac{{\sqrt[3]{{12}} \cdot \sqrt[3]{6}}}{{\sqrt[3]{{576}}}} - \frac{{\sqrt[3]{{32}}}}{{\sqrt[3]{4}}} = \sqrt[3]{{\frac{{12 \cdot 6}}{{576}}}} - \sqrt[3]{{\frac{{32}}{4}}} = \sqrt[3]{{\frac{1}{8}}} - \sqrt[3]{8} = \frac{1}{2} - 2 =  - \frac{3}{2}\).

a) \(\sqrt[3]{{\frac{1}{2}}} \cdot \sqrt[3]{{ - 18}} \cdot \sqrt[3]{3} = \sqrt[3]{{\frac{1}{2} \cdot ( - 18) \cdot 3}} = \sqrt[3]{{ - 27}} =  - 3\).

b) \(\sqrt[3]{{\left( {\sqrt 2  + 1} \right)\left( {3 + 2\sqrt 2 } \right)}} = \sqrt[3]{{\left( {\sqrt 2  + 1} \right){{\left( {\sqrt 2  + 1} \right)}^2}}} = \sqrt[3]{{{{\left( {\sqrt 2  + 1} \right)}^3}}} = \sqrt 2  + 1\).

\(\sqrt[3]{{\left( {4 - 2\sqrt 3 } \right)\left( {\sqrt 3  - 1} \right)}} = \sqrt[3]{{{{\left( {\sqrt 3  - 1} \right)}^2}\left( {\sqrt 3  - 1} \right)}} = \sqrt[3]{{{{\left( {\sqrt 3  - 1} \right)}^3}}} = \sqrt 3  - 1\).

c) \(\left( {\frac{1}{2}\sqrt[3]{2} - \frac{1}{4}\sqrt[3]{{16}}} \right) \cdot \sqrt[3]{4} = \left( {\frac{1}{2}\sqrt[3]{2} - \frac{1}{2}\sqrt[3]{2}} \right) \cdot \sqrt[3]{4} = 0\).

d) \[\left( {\frac{1}{2}\sqrt[3]{9} - 2\sqrt[3]{3} + 3\sqrt[3]{{\frac{1}{3}}}} \right):2\sqrt[3]{{\frac{1}{3}}} = \left( {\frac{1}{2}\sqrt[3]{9} - 2\sqrt[3]{3} + \sqrt[3]{9}} \right):\frac{2}{3}\sqrt[3]{9} = \left( {\frac{3}{2}\sqrt[3]{9} - 2\sqrt[3]{3}} \right):\frac{2}{3}\sqrt[3]{9} = \frac{9}{4} - \sqrt[3]{9}\].

e) \(\left( {\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}} \right)\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right) = \left( {\sqrt[3]{{{3^2}}} - \sqrt[3]{{3 \cdot 2}} + \sqrt[3]{{{2^2}}}} \right)\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right) = {\left( {\sqrt[3]{3}} \right)^3} + {\left( {\sqrt[3]{2}} \right)^3} = 3 + 2 = 5\).

a) \[\sqrt[3]{{ - 64}} - \sqrt[3]{{125}} + \sqrt[3]{{216}} =  - 4 - 5 + 6 =  - 3\]

b) Ta có \[{\left( {\sqrt[3]{4} + 1} \right)^3} - {\left( {\sqrt[3]{4} - 1} \right)^3}\]

\[ = \left( {\sqrt[3]{4} + 1 - \sqrt[3]{4} + 1} \right)\left[ {{{\left( {\sqrt[3]{4} + 1} \right)}^2} + \left( {\sqrt[3]{4} + 1} \right)\left( {\sqrt[3]{4} - 1} \right) + {{\left( {\sqrt[3]{4} - 1} \right)}^2}} \right]\]

\[ = 2.\left( {\sqrt[3]{{16}} + 2\sqrt[3]{4} + 1 + \sqrt[3]{{16}} - 1 + \sqrt[3]{{16}} - 2\sqrt[3]{4} + 1} \right) = 2\left( {3\sqrt[3]{{16}} + 1} \right) = 2\left( {6\sqrt[3]{2} + 1} \right)\]

c) Ta có \[\left( {12\sqrt[3]{2} + \sqrt[3]{{16}} - 2\sqrt[3]{2}} \right)\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right)\]

\[ = \left( {12\sqrt[3]{2} + 2\sqrt[3]{2} - 2\sqrt[3]{2}} \right)\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right) = 12\sqrt[3]{2}\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right) = 120 - 36 = 84\]

Ta có \[x = \frac{2}{{2\sqrt[3]{2} + 2 + \sqrt[3]{4}}} = \frac{2}{{\sqrt[3]{{16}} + 2 + \sqrt[3]{4}}}\]

\[ = \frac{2}{{{{\left( {\sqrt[3]{4}} \right)}^2} + \sqrt[3]{2}.\sqrt[3]{4} + {{\left( {\sqrt[3]{2}} \right)}^2}}}\]

\[ = \frac{{2\left( {\sqrt[3]{4} - \sqrt[3]{2}} \right)}}{{4 - 2}} = \sqrt[3]{4} - \sqrt[3]{2}\]

Tương tự \(y = \frac{6}{{2\sqrt[3]{2} - 2 + \sqrt[3]{4}}} = \sqrt[3]{4} + \sqrt[3]{2}\)

Do đó: \(x{y^3} - {x^3}y = xy\left( {{y^2} - {x^2}} \right) = xy\left( {y - x} \right)\left( {y + x} \right) = 8\left( {2\sqrt[3]{2} - \sqrt[3]{4}} \right)\)

a) \(A = \sqrt[3]{{{2^3} + 3 \cdot {2^2}\left( {\sqrt 2 } \right) + 3 \cdot 2 \cdot {{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^3}}} + \sqrt[3]{{{2^3} - 3 \cdot {2^2}\left( {\sqrt 2 } \right) + 3 \cdot 2 \cdot {{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^3}}}\)

\( = \sqrt[3]{{{{\left( {2 + \sqrt 2 } \right)}^3}}} + \sqrt[3]{{{{\left( {2 - \sqrt 2 } \right)}^3}}} = 2 + \sqrt 2  + 2 - \sqrt 2  = 4\)

b) \({B^3} = 182 + \sqrt {33125}  + 182 - \sqrt {33125}  + 3\sqrt[3]{{{{182}^2} - 33125}}B = 364 - 3B.\)

Khi đó  \({B^3} + 3B - 364 = 0\) nên \(\left( {B - 7} \right)\left( {{B^2} + 7B + 52} \right) = 0\), suy ra \(B = 7\).

(do \({B^2} + 7B + 52 = {\left( {B + \frac{7}{2}} \right)^2} + \frac{{159}}{4} > 0\)).