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Câu hỏi: Cho ${{\log }_{3}}5=a,{{\log }_{3}}6=b,{{\log }_{3}}22=c.$ Tính $P={{\log }_{3}}\dfrac{90}{11}$ theo $a,b,c$.
A. $P=2a+b-c.$
B. $P=a+2b-c.$
C. $P=2a+b+c.$
D. $P=2a-b+c.$
A. $P=2a+b-c.$
B. $P=a+2b-c.$
C. $P=2a+b+c.$
D. $P=2a-b+c.$
Ta có
$P={{\log }_{3}}\dfrac{180}{22}={{\log }_{3}}180-{{\log }_{3}}22={{\log }_{3}}\left( {{5.6}^{2}} \right)-c={{\log }_{3}}5+2{{\log }_{3}}6-c=a+2b-c.$
$P={{\log }_{3}}\dfrac{180}{22}={{\log }_{3}}180-{{\log }_{3}}22={{\log }_{3}}\left( {{5.6}^{2}} \right)-c={{\log }_{3}}5+2{{\log }_{3}}6-c=a+2b-c.$
Đáp án B.