Bài tập Thực hiện phép tính và rút gọn biểu thức lớp 9 (có lời giải)

Ta có: \[\sqrt[3]{{512}} = \sqrt[3]{{{8^3}}} = 8.\]

\[\sqrt[3]{{ - 729}} = \sqrt[3]{{ - {9^3}}} =  - 9.\]

\[\sqrt[3]{{0,064}} = \sqrt[3]{{{{(0,4)}^3}}} = 0,4.\]

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

\[\sqrt[3]{{ - 0,008}} = \sqrt[3]{{{{( - 0,2)}^3}}} =  - 0,2.\]

a). \[\sqrt[3]{{27}} - \sqrt[3]{{ - 8}} - \sqrt[3]{{125}} = \sqrt[3]{{{3^3}}} - \sqrt[3]{{{2^3}}} - \sqrt[3]{{{5^3}}} = 3 + 2 - 5 = 0\]

b). \[\frac{{\sqrt[3]{{135}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{54}}.\sqrt[3]{4}. = \sqrt[3]{{\frac{{135}}{5}}} - \sqrt[3]{{54.4}}. = \sqrt[3]{{27}} - \sqrt[3]{{216}} = 3 - 6 =  - 3\]

a). Ta có:\[A = \left( {\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}} \right)\left( {\sqrt[3]{3} - \sqrt[3]{2}} \right) = \left( {{{\left( {\sqrt[3]{3}} \right)}^2} + \sqrt[3]{3}\sqrt[3]{2} + {{\left( {\sqrt[3]{2}} \right)}^2}} \right)\left( {\sqrt[3]{3} - \sqrt[3]{2}} \right)\]

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

b). Áp dụng hằng đẳng thức \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b)\]

Ta có: \[{B^3} = {\left( {\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}} \right)^3} = 2 + \sqrt 5  + 2 - \sqrt 5  + 3\sqrt[3]{{2 + \sqrt 5 }}\sqrt[3]{{2 - \sqrt 5 }}\left( {\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}} \right)\]

\[ = 4 + 3\sqrt[3]{{\left( {2 + \sqrt 5 } \right)\left( {2 - \sqrt 5 } \right)}}B = 4 + 3\sqrt[3]{{4 - 5}}B = 4 - 3B\]

\[\begin{array}{l}{B^3} + 3B - 4 = 0\\{B^3} - 1 + 3B - 3 = 0\end{array}\]

\[\left( {B - 1} \right)\left( {{B^2} + B + 4} \right) = 0\]

\[B = 1\, \left( {{B^2} + B + 4 = {B^2} + 2\frac{1}{2}B + \frac{1}{4} + \frac{{15}}{4} = {{\left( {B + \frac{1}{2}} \right)}^2} + \frac{{15}}{4} > 0} \right)\]

Vậy \[B = \sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }} = 1.\]

a) Ta có \(\sqrt[3]{{{x^3} + 1 + 3x\left( {x + 1} \right)}} = \sqrt[3]{{{{(x + 1)}^3}}} = x + 1\).    

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