• Bài tập tự luyện dạng 2

Câu 1:

Tính giá trị của biểu thức sau

a) \[{{\left( \sqrt[3]{3}+\sqrt[3]{2} \right)}^{3}}\] b) \[\left( \sqrt[3]{5}-\sqrt[3]{3} \right).\left( \sqrt[3]{25}+\sqrt[3]{15}+\sqrt[3]{9} \right)\]

c) \[\sqrt[3]{162}.\sqrt[3]{-2}.\sqrt[3]{\frac{2}{3}}\] d) \[\sqrt[3]{2}:\sqrt[3]{16}-\sqrt[3]{22\frac{1}{2}}:\sqrt[3]{53\frac{1}{3}}\]

Câu 2:

Rút gọn biểu thức:

a) \[\sqrt[3]{3}.(5\sqrt[3]{18}-\sqrt[3]{144})+\sqrt[3]{5}.\sqrt[3]{50}\] b) \[\left( 12.\sqrt[3]{2}+\sqrt[3]{16}-2.\sqrt[3]{2} \right).\left( 5.\sqrt[3]{4}-3.\sqrt[3]{\frac{1}{2}} \right)\]

Câu 3:

Tính

a) \[A=\frac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}\] b) \[B=\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}\]

c) \[C=\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}\]

Câu 4:

Thực hiện phép tính sau

a) \[A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\] b) \[A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\]

c) \[C=\left( 2-\sqrt{3} \right).\sqrt[3]{26+15\sqrt{3}}\]

LỜI GIẢI BÀI TẬP TỰ LUYỆN

Câu 1:

a) \[5+3\sqrt[3]{6}.\left( \sqrt[3]{3}+\sqrt[3]{2} \right).\] b) \[2.\] c) \[-6.\] d) \[-\frac{1}{4}.\]

Câu 2:

a) \[14\sqrt[3]{2}.\] b) \[84.\]

Câu 3:

a) \[A=\frac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\frac{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{8}}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\frac{\sqrt[3]{2}.\left( \sqrt[3]{4}+\sqrt[3]{2}+1 \right)}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\sqrt[3]{2}.\]

b) \[B=\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}.\] Ta có : \[\sqrt[3]{10+6\sqrt{3}}=\sqrt{3}+1\Rightarrow B=\sqrt{3}+1.\]

c) \[C=\frac{4+2\sqrt{3}}{\sqrt{3}+1}=\frac{{{\left( \sqrt{3}+1 \right)}^{2}}}{\sqrt{3}+1}=\sqrt{3}+1\].

Câu 4:

a) \[A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\]

\[=\sqrt[3]{{{\left( \frac{1+\sqrt{5}}{2} \right)}^{3}}}+\sqrt[3]{{{\left( \frac{1-\sqrt{5}}{2} \right)}^{3}}}\]

\[=\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}\]

\[=1.\]

b) \[B=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\]

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

\[=\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\]

\[=3.\]

c) Ta có \[26+15\sqrt{3}={{\left( 2+\sqrt{3} \right)}^{3}}\]nên \[C=\left( 2-\sqrt{3} \right).\sqrt[3]{{{\left( 2+\sqrt{3} \right)}^{3}}}=\left( 2-\sqrt{3} \right)\left( 2+\sqrt{3} \right)=1.\]