• Bài tập tự luyện dạng 3

Câu 1:

Cho \[x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}};y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\].Tính giá trị biểu thức \[P=\frac{xy}{x+y}\]

Câu 2:

Cho \[x=\frac{\sqrt[3]{8-3\sqrt{5}}+\sqrt[3]{64-12\sqrt{20}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}};y=\frac{\sqrt[3]{9}-\sqrt{2}}{\sqrt[3]{3}+\sqrt[4]{2}}+\frac{\sqrt{2}-9\sqrt[3]{9}}{\sqrt[4]{2}-\sqrt[3]{81}}\]

Tính giá trị của biểu thức \[Q=x.y.\]

Câu 3:

Rút gọn biểu thức \[P=\left( \frac{x\sqrt[3]{x}-2x\sqrt[3]{y}+\sqrt[3]{{{x}^{2}}{{y}^{2}}}}{\sqrt[3]{{{x}^{2}}}-\sqrt[3]{xy}}+\frac{\sqrt[3]{{{x}^{2}}y}-\sqrt[3]{x{{y}^{2}}}}{\sqrt[3]{x}-\sqrt[3]{y}} \right).\frac{1}{\sqrt[3]{{{x}^{2}}}}\]

Câu 4:

Chứng minh rằng, nếu \[a{{x}^{3}}=b{{y}^{3}}=c{{z}^{3}}\]và \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\] thì \[\sqrt[3]{a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}.\]

Câu 5:

Chứng minh đẳng thức

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

LỜI GIẢI BÀI TẬP TỰ LUYỆN

Câu 1:

Ta có

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

\[y=\frac{6.\left( \sqrt[3]{4}+\sqrt[3]{2} \right)}{\left( \sqrt[3]{4}+\sqrt[3]{2} \right).\left( 2\sqrt[3]{2}-2+\sqrt[3]{4} \right)}=\frac{6.\left( \sqrt[3]{4}+\sqrt[3]{2} \right)}{{{\left( \sqrt[3]{4} \right)}^{3}}+{{\left( \sqrt[3]{2} \right)}^{3}}}=\sqrt[3]{4}+\sqrt[3]{2}.\]

Suy ra \[P=\frac{x.y}{x+y}=\frac{\left( \sqrt[3]{4}-\sqrt[3]{2} \right).\left( \sqrt[3]{4}+\sqrt[3]{2} \right)}{\sqrt[3]{4}-\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{2}}=\frac{\left[ {{\left( \sqrt[3]{4} \right)}^{2}}-{{\left( \sqrt[3]{2} \right)}^{2}} \right]}{2.\sqrt[3]{4}}=\frac{\sqrt[3]{4}-1}{2}.\]

Câu 2:

Ta có

\[x=\frac{\sqrt[3]{8-3\sqrt{5}}+\sqrt[3]{64-12\sqrt{20}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\frac{\sqrt[3]{8-3\sqrt{5}}+2\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}\]

\[=\frac{3\sqrt[3]{\left( 8-3\sqrt{5} \right)\left( 8+3\sqrt{5} \right)}}{\sqrt[3]{57}}=\frac{3\sqrt[3]{19}}{\sqrt[3]{57}}=\sqrt[3]{9}.\]

\[y=\frac{\sqrt[3]{9}-\sqrt{2}}{\sqrt[3]{3}+\sqrt[4]{2}}+\frac{\sqrt{2}-9\sqrt[3]{9}}{\sqrt[4]{2}-\sqrt[3]{81}}=\frac{\left( \sqrt[3]{3}-\sqrt[4]{2} \right)\left( \sqrt[3]{3}+\sqrt[4]{2} \right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\frac{\left( \sqrt[4]{2}-3\sqrt[3]{3} \right)\left( \sqrt[4]{2}+3\sqrt[3]{3} \right)}{\sqrt[4]{2}-3\sqrt[3]{3}}\]

\[=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+3\sqrt[3]{3}=4\sqrt[3]{3}\].

\[Q=x.y=12.\]

Câu 3:

Đặt \[\sqrt[3]{x}=a;\sqrt[3]{y}=b.\]

\[P=\left( \frac{{{a}^{4}}-2{{a}^{3}}b+{{a}^{2}}{{b}^{2}}}{{{a}^{2}}-ab}+\frac{{{a}^{2}}b-a{{b}^{2}}}{a-b} \right).\frac{1}{{{a}^{2}}}\]

\[=\left( \frac{{{\left( {{a}^{2}}-ab \right)}^{2}}}{{{a}^{2}}-ab}+\frac{ab\left( a-b \right)}{a-b} \right).\frac{1}{{{a}^{2}}}\]

\[=\left( {{a}^{2}}-ab+ab \right).\frac{1}{{{a}^{2}}}=1.\]

Vậy \[P=1.\]

Câu 4:

Đặt \[a{{x}^{3}}=b{{y}^{3}}=c{{z}^{3}}=t\Rightarrow a=\frac{t}{{{x}^{3}}};b=\frac{t}{{{y}^{3}}};c=\frac{t}{{{z}^{3}}}.\]

Ta có \[VT=\sqrt[3]{a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}}=\sqrt[3]{\frac{t}{{{x}^{3}}}.{{x}^{2}}+\frac{t}{{{y}^{3}}}.{{y}^{2}}+\frac{t}{{{z}^{3}}}.{{z}^{2}}}=\sqrt[3]{t.\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)}=\sqrt[3]{t}.\]

\[VP=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{\frac{t}{{{x}^{3}}}}+\sqrt[3]{\frac{t}{{{y}^{3}}}}+\sqrt[3]{\frac{t}{{{z}^{3}}}}=\frac{\sqrt[3]{t}}{x}+\frac{\sqrt[3]{t}}{y}+\frac{\sqrt[3]{t}}{z}=\sqrt[3]{t}.\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)=\sqrt[3]{t}.\]

Vậy VT=VP (đẳng thức được chứng minh).

Câu 5:

\[VP=\frac{1}{2}.\left( \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z} \right).\left( \sqrt[3]{{{x}^{2}}}-2.\sqrt[3]{xy}+\sqrt[3]{{{y}^{2}}}+\sqrt[3]{{{y}^{2}}}-2.\sqrt[3]{yz}+\sqrt[3]{z}+\sqrt[3]{z}-2.\sqrt[3]{zx}+\sqrt[3]{{{x}^{2}}} \right)\]

\[=\frac{1}{2}.\left( \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z} \right).\left( 2\sqrt[3]{{{x}^{2}}}+2\sqrt[3]{{{y}^{2}}}+2\sqrt[3]{z}-2\sqrt[3]{xy}-2\sqrt[3]{yz}-2\sqrt[3]{zx} \right)\]

\[=\frac{1}{2}.\left( 2\sqrt[3]{{{x}^{2}}}+2\sqrt[3]{x{{y}^{2}}}+2\sqrt[3]{xz}-2\sqrt[3]{{{x}^{2}}y}-2\sqrt[3]{xyz}-2\sqrt[3]{{{x}^{2}}z}+2\sqrt[3]{{{x}^{2}}y}+2\sqrt[3]{{{y}^{3}}}+2\sqrt[3]{yz}-2\sqrt[3]{x{{y}^{2}}}-2\sqrt[3]{{{y}^{2}}z} \right)\]

\[-2\sqrt[3]{xyz}+\sqrt[3]{{{x}^{2}}z}+2\sqrt[3]{z{{y}^{2}}}+2\sqrt[3]{{{z}^{3}}}-2\sqrt[3]{xyz}-2\sqrt[3]{y{{z}^{2}}}-2\sqrt[3]{x{{z}^{2}}})\]

\[=x+y+z-3\sqrt[3]{xyz}=VT.\] Vậy đẳng thức được chứng minh.