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Câu hỏi: Cho $lo{{g}_{3}}5=a,lo{{g}_{3}}6=b,lo{{g}_{3}}22=c$ Tính $P={{\log }_{3}}\left( \dfrac{90}{11} \right)$ theo a, b, _C_.
A. $P=2a+b-c$
B. $P=a+2b-c$
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. _A_. $P=2a+b-c$
_B_. $P=a+2b-c$
_C_. $P=2a+b+c$
D. $P=-2a-b+c$
Lời giải
Ta có: $P=lo{{g}_{3}}\left( \dfrac{90}{11} \right)=lo{{g}_{3}}\left( \dfrac{180}{22} \right)=lo{{g}_{3}}180-lo{{g}_{3}}22=lo{{g}_{3}}\left( 36.5 \right)-lo{{g}_{3}}36+lo{{g}_{3}}5-lo{{g}_{3}}22$
$=lo{{g}_{3}}\left( {{6}^{2}} \right)+lo{{g}_{3}}5-lo{{g}_{3}}22=2lo{{g}_{3}}6+lo{{g}_{3}}5-lo{{g}_{3}}22=a+2b-c$
Vậy $P=a+2b-c.$
Ta có: $P=lo{{g}_{3}}\left( \dfrac{90}{11} \right)=lo{{g}_{3}}\left( \dfrac{180}{22} \right)=lo{{g}_{3}}180-lo{{g}_{3}}22=lo{{g}_{3}}\left( 36.5 \right)-lo{{g}_{3}}36+lo{{g}_{3}}5-lo{{g}_{3}}22$
$=lo{{g}_{3}}\left( {{6}^{2}} \right)+lo{{g}_{3}}5-lo{{g}_{3}}22=2lo{{g}_{3}}6+lo{{g}_{3}}5-lo{{g}_{3}}22=a+2b-c$
Vậy $P=a+2b-c.$
Đáp án B.