Google
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. $T=\dfrac{2}{9}$.
B. $T=\dfrac{8}{9}$.
C. $T=\dfrac{3}{2}$.
D. $T=\dfrac{2}{3}$.
A. $T=\dfrac{2}{9}$.
B. $T=\dfrac{8}{9}$.
C. $T=\dfrac{3}{2}$.
D. $T=\dfrac{2}{3}$.
Ta có ${{\log }_{\sqrt{8}}}\sqrt[3]{a{{b}^{2}}}={{\log }_{{{2}^{\dfrac{3}{2}}}}}{{(a{{b}^{2}})}^{\dfrac{1}{3}}}=\dfrac{2}{3}\left( \dfrac{1}{3}{{\log }_{2}}a+\dfrac{2}{3}{{\log }_{2}}b \right)=\dfrac{2}{3}{{\log }_{2}}a+\dfrac{4}{9}{{\log }_{2}}b$.
Với ${{\log }_{2}}a=x,{{\log }_{2}}b=y$ ta suy ra $m=\dfrac{2}{9};n=\dfrac{4}{9}$.
Vậy $T=\dfrac{2}{9}+\dfrac{4}{9}=\dfrac{2}{3}$ .
Với ${{\log }_{2}}a=x,{{\log }_{2}}b=y$ ta suy ra $m=\dfrac{2}{9};n=\dfrac{4}{9}$.
Vậy $T=\dfrac{2}{9}+\dfrac{4}{9}=\dfrac{2}{3}$ .
Đáp án D.