Câu hỏi: Cho ${{\log }_{\dfrac{1}{2}}}\left( \dfrac{1}{5} \right)=a$. Khẳng định nào sau đây đúng?
A. ${{\log }_{2}}25+{{\log }_{2}}\sqrt{5}=\dfrac{5a}{2}.$
B. ${{\log }_{2}}5=-a.$
C. ${{\log }_{5}}4=-\dfrac{2}{a}.$
D. ${{\log }_{2}}\dfrac{1}{5}+{{\log }_{2}}\dfrac{1}{25}=3a.$
A. ${{\log }_{2}}25+{{\log }_{2}}\sqrt{5}=\dfrac{5a}{2}.$
B. ${{\log }_{2}}5=-a.$
C. ${{\log }_{5}}4=-\dfrac{2}{a}.$
D. ${{\log }_{2}}\dfrac{1}{5}+{{\log }_{2}}\dfrac{1}{25}=3a.$
Đáp án B sai vì theo giả thiết ${{\log }_{\dfrac{1}{2}}}\left( \dfrac{1}{5} \right)=a\Leftrightarrow {{\log }_{{{2}^{-1}}}}\left( {{5}^{-1}} \right)=a\Leftrightarrow {{\log }_{2}}5=a$.
Đáp án C sai vì ${{\log }_{5}}4={{\log }_{5}}{{2}^{2}}=2{{\log }_{5}}2=\dfrac{2}{{{\log }_{2}}5}=\dfrac{2}{a}$.
Đáp án D sai vì ${{\log }_{2}}\dfrac{1}{5}+{{\log }_{2}}\dfrac{1}{25}={{\log }_{2}}{{5}^{-1}}+{{\log }_{2}}{{5}^{-2}}=-{{\log }_{2}}5-2{{\log }_{2}}5=-3a$.
Đáp án A đúng vì ${{\log }_{2}}25+{{\log }_{2}}\sqrt{5}={{\log }_{2}}{{5}^{2}}+{{\log }_{2}}{{5}^{\dfrac{1}{2}}}=2{{\log }_{2}}5+\dfrac{1}{2}{{\log }_{2}}5=\dfrac{5a}{2}$.
Đáp án C sai vì ${{\log }_{5}}4={{\log }_{5}}{{2}^{2}}=2{{\log }_{5}}2=\dfrac{2}{{{\log }_{2}}5}=\dfrac{2}{a}$.
Đáp án D sai vì ${{\log }_{2}}\dfrac{1}{5}+{{\log }_{2}}\dfrac{1}{25}={{\log }_{2}}{{5}^{-1}}+{{\log }_{2}}{{5}^{-2}}=-{{\log }_{2}}5-2{{\log }_{2}}5=-3a$.
Đáp án A đúng vì ${{\log }_{2}}25+{{\log }_{2}}\sqrt{5}={{\log }_{2}}{{5}^{2}}+{{\log }_{2}}{{5}^{\dfrac{1}{2}}}=2{{\log }_{2}}5+\dfrac{1}{2}{{\log }_{2}}5=\dfrac{5a}{2}$.
Đáp án A.