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Câu hỏi: Cho hai số thực dương $a,b$. Rút gọn biểu thức $A=\dfrac{{{a}^{\dfrac{1}{3}}}\sqrt{b}+{{b}^{\dfrac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}={{a}^{m}}.{{b}^{n}}$. Tổng của $m+n$ là
A. $\dfrac{5}{6}$
B. $\dfrac{1}{6}$
C. $\dfrac{1}{9}$
D. $\dfrac{2}{3}$
A. $\dfrac{5}{6}$
B. $\dfrac{1}{6}$
C. $\dfrac{1}{9}$
D. $\dfrac{2}{3}$
Ta có: $A=\dfrac{{{a}^{\dfrac{1}{3}}}\sqrt{b}+{{b}^{\dfrac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}=\dfrac{{{a}^{\dfrac{1}{3}}}.{{b}^{\dfrac{1}{2}}}+{{b}^{\dfrac{1}{3}}}.{{a}^{\dfrac{1}{2}}}}{{{a}^{\dfrac{1}{6}}}+{{b}^{\dfrac{1}{6}}}}=\dfrac{{{a}^{\dfrac{1}{3}}}.{{b}^{\dfrac{1}{3}}}\left( {{b}^{\dfrac{1}{6}}}+{{a}^{\dfrac{1}{6}}} \right)}{{{a}^{\dfrac{1}{6}}}+{{b}^{\dfrac{1}{6}}}}={{a}^{\dfrac{1}{3}}}.{{b}^{\dfrac{1}{3}}}$
$\Rightarrow m+n=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}$
$\Rightarrow m+n=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}$
Đáp án D.