• Bài tập tự luyện dạng 1

Câu 1:

Tính

a) \[\sqrt[3]{\frac{3}{4}}.\sqrt[3]{\frac{9}{16}}\] b) \[\frac{\sqrt[3]{54}}{\sqrt[3]{-2}}\] c) \[\sqrt[3]{8}-\frac{1}{3}\sqrt[3]{-27}+\frac{2}{5}\sqrt[3]{125}\]

d) \[\sqrt[3]{1000}-\sqrt[3]{512}-2\sqrt[3]{-64}\] e) \[\sqrt[3]{-1,25}.\sqrt[3]{0,5}.\sqrt[3]{\frac{16}{10}}\]

Câu 2:

Rút gọn

a) \[\sqrt[3]{-8{{a}^{6}}}\] b) \[\sqrt[3]{-0,512{{x}^{9}}}\]

c) \[\sqrt[3]{-\frac{{{y}^{6}}}{729}}\] d) \[\sqrt[3]{\frac{64}{343{{b}^{6}}}}\]với \[b\ne 0\]

e) \[\sqrt[3]{\frac{1331{{x}^{3}}{{y}^{6}}}{8{{x}^{9}}}}\]với \[x\ne 0\] f) \[\sqrt[3]{\frac{-27{{x}^{2}}{{y}^{4}}}{{{x}^{5}}y}}\]với \[x\ne 0,y\ne 0\]

LỜI GIẢI BÀI TẬP TỰ LUYỆN

Câu 1:

a) Ta có \[\sqrt[3]{\frac{3}{4}}.\sqrt[3]{\frac{9}{16}}=\sqrt[3]{\frac{3}{4}.\frac{9}{16}}=\sqrt[3]{\frac{27}{64}}=\frac{\sqrt[3]{27}}{\sqrt[3]{64}}=\frac{3}{4}.\]

b) Ta có \[\frac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\frac{-54}{2}}=\sqrt[3]{-27}=\sqrt[3]{{{\left( -3 \right)}^{3}}}=-3.\]

c) Ta có \[\sqrt[3]{8}-\frac{1}{3}\sqrt[3]{-27}+\frac{2}{5}\sqrt[3]{125}=2-\frac{1}{3}.\left( -3 \right)+\frac{2}{5}.5=2+1+2=5.\]

d) Ta có \[\sqrt[3]{1000}-\sqrt[3]{512}-2\sqrt[3]{-64}=\sqrt[3]{{{10}^{3}}}-\sqrt[3]{{{8}^{3}}}-2\sqrt[3]{{{\left( -4 \right)}^{3}}}=10-8-2.\left( -4 \right)=10.\]

e) Ta có \[\sqrt[3]{-1,25}.\sqrt[3]{0,5}.\sqrt[3]{\frac{16}{10}}=\sqrt[3]{-1,25.0,5.\frac{16}{10}}=\sqrt[3]{-1}=-1.\]

Câu 2:

a) Ta có \[\sqrt[3]{-8{{a}^{6}}}=\sqrt[3]{{{\left( -2{{a}^{2}} \right)}^{3}}}=-2{{a}^{2}}.\]

b) Ta có \[\sqrt[3]{-0,512{{x}^{9}}}=\sqrt[3]{\frac{-512{{x}^{9}}}{1000}}=\sqrt[3]{\frac{-64{{x}^{9}}}{125}}=\sqrt[3]{{{\left( \frac{-4{{x}^{3}}}{5} \right)}^{3}}}=\frac{-4{{x}^{3}}}{5}.\]

c) Ta có \[\sqrt[3]{-\frac{{{y}^{6}}}{729}}=\sqrt[3]{{{\left( -\frac{{{y}^{2}}}{9} \right)}^{3}}}=\frac{-{{y}^{2}}}{9}.\]

d) Ta có \[\sqrt[3]{\frac{64}{343{{b}^{6}}}}=\sqrt[3]{{{\left( \frac{4}{7{{b}^{2}}} \right)}^{3}}}=\frac{4}{7{{b}^{2}}}.\]

e) Ta có \[\sqrt[3]{\frac{1331{{x}^{3}}{{y}^{6}}}{8{{x}^{9}}}}=\sqrt[3]{\frac{1331{{y}^{6}}}{8{{x}^{6}}}}=\sqrt[3]{{{\left( \frac{11{{y}^{2}}}{2{{x}^{2}}} \right)}^{3}}}=\frac{11{{y}^{2}}}{2{{x}^{2}}}.\]

f) Ta có \[\sqrt[3]{\frac{-27{{x}^{2}}{{y}^{4}}}{{{x}^{5}}y}}=\sqrt[3]{\frac{-27{{y}^{3}}}{{{x}^{3}}}}=\sqrt[3]{{{\left( \frac{-3y}{x} \right)}^{3}}}=\frac{-3y}{x}.\]