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Câu hỏi: Đặt ${{\log }_{2}}a=x,{{\log }_{2}}b=y$. Biết ${{\log }_{\sqrt{8}}}\sqrt[3]{a{{b}^{2}}}=mx+ny$. Tìm $T=m+n$.
A. $T=\dfrac{3}{2}$.
B. $T=\dfrac{2}{3}$.
C. $T=\dfrac{2}{9}$.
D. $T=\dfrac{8}{9}$.
A. $T=\dfrac{3}{2}$.
B. $T=\dfrac{2}{3}$.
C. $T=\dfrac{2}{9}$.
D. $T=\dfrac{8}{9}$.
Ta có $mx+ny={{\log }_{{{2}^{\dfrac{3}{2}}}}}{{\left( a{{b}^{2}} \right)}^{\dfrac{1}{3}}}=\dfrac{\dfrac{1}{3}}{\dfrac{3}{2}}{{\log }_{2}}\left( a{{b}^{2}} \right)=\dfrac{2}{9}\left( {{\log }_{2}}a+{{\log }_{2}}{{b}^{2}} \right)=\dfrac{2}{9}\left( {{\log }_{2}}a+2{{\log }_{2}}b \right)$
$\Rightarrow m=\dfrac{2}{9};n=\dfrac{4}{9}\Rightarrow m+n=\dfrac{2}{3}$.
$\Rightarrow m=\dfrac{2}{9};n=\dfrac{4}{9}\Rightarrow m+n=\dfrac{2}{3}$.
Đáp án B.