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Câu hỏi: . Cho ${{\log }_{3}}a=5$ và ${{\log }_{3}}b=\dfrac{2}{3}$. Tính giá trị của biểu thức $I=2{{\log }_{6}}\left[ {{\log }_{5}}\left( 5a \right) \right]+{{\log }_{\dfrac{1}{9}}}{{b}^{3}}$.
A. $I=3$
B. $I=-2$
C. $I=1$
D. $I={{\log }_{6}}5+1$
A. $I=3$
B. $I=-2$
C. $I=1$
D. $I={{\log }_{6}}5+1$
Sử dụng các công thức:
${{\log }_{a}}f\left( x \right)+{{\log }_{a}}g\left( x \right)={{\log }_{a}}\left[ f\left( x \right)g\left( x \right) \right]\left( 0<a\ne 1,f\left( x \right)>0,g\left( x \right)>0 \right)$
${{\log }_{{{a}^{n}}}}{{b}^{m}}=\dfrac{m}{n}{{\log }_{a}}b\left( 0<a\ne 1,b>0 \right)$
$I=2{{\log }_{6}}\left[ {{\log }_{5}}\left( 5\text{a} \right) \right]+{{\log }_{\dfrac{1}{9}}}{{b}^{3}}=2{{\log }_{6}}\left[ 1+{{\log }_{5}}a \right]-\dfrac{3}{2}{{\log }_{3}}b=2{{\log }_{6}}6-\dfrac{3}{2}.\dfrac{2}{3}=2.1-1=1$.
${{\log }_{a}}f\left( x \right)+{{\log }_{a}}g\left( x \right)={{\log }_{a}}\left[ f\left( x \right)g\left( x \right) \right]\left( 0<a\ne 1,f\left( x \right)>0,g\left( x \right)>0 \right)$
${{\log }_{{{a}^{n}}}}{{b}^{m}}=\dfrac{m}{n}{{\log }_{a}}b\left( 0<a\ne 1,b>0 \right)$
$I=2{{\log }_{6}}\left[ {{\log }_{5}}\left( 5\text{a} \right) \right]+{{\log }_{\dfrac{1}{9}}}{{b}^{3}}=2{{\log }_{6}}\left[ 1+{{\log }_{5}}a \right]-\dfrac{3}{2}{{\log }_{3}}b=2{{\log }_{6}}6-\dfrac{3}{2}.\dfrac{2}{3}=2.1-1=1$.
Đáp án C.