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Câu hỏi: Cho hai số thực dương $a$ và $b.$ Rút gọn biểu thức $A=\dfrac{{{a}^{\dfrac{1}{3}}}\sqrt{b}+{{b}^{\dfrac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}.$
A. $A=\sqrt[3]{ab}$
B. $A=\sqrt[6]{ab}$
C. $\dfrac{1}{\sqrt[3]{ab}}$
D. $\dfrac{1}{\sqrt[6]{ab}}$
A. $A=\sqrt[3]{ab}$
B. $A=\sqrt[6]{ab}$
C. $\dfrac{1}{\sqrt[3]{ab}}$
D. $\dfrac{1}{\sqrt[6]{ab}}$
Ta có: $A=\dfrac{{{a}^{\dfrac{1}{3}}}\sqrt{b}+{{b}^{\dfrac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}=\dfrac{{{a}^{\dfrac{1}{3}}}{{b}^{\dfrac{1}{2}}}+{{b}^{\dfrac{1}{3}}}{{a}^{\dfrac{1}{2}}}}{{{a}^{\dfrac{1}{6}}}+{{b}^{\dfrac{1}{6}}}}=\dfrac{{{a}^{\dfrac{1}{3}}}{{b}^{\dfrac{1}{3}}}\left( {{b}^{\dfrac{1}{6}}}+{{a}^{\dfrac{1}{6}}} \right)}{{{a}^{\dfrac{1}{6}}}+{{b}^{\dfrac{1}{6}}}}={{\left( ab \right)}^{\dfrac{1}{3}}}=\sqrt[3]{ab}.$
Đáp án A.