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Câu hỏi: Đặt $a={{\log }_{2}}3,b={{\log }_{5}}3.$ Khi đó ${{\log }_{6}}45$ được biểu diễn theo a và b là
A. ${{\log }_{6}}45=\dfrac{a+2\text{a}b}{ab}.$
B. ${{\log }_{6}}45=\dfrac{2a-2\text{a}b}{ab}.$
C. ${{\log }_{6}}45=\dfrac{a+2\text{a}b}{ab+b}.$
D. ${{\log }_{6}}45=\dfrac{2{{\text{a}}^{2}}-2\text{a}b}{ab+b}.$
A. ${{\log }_{6}}45=\dfrac{a+2\text{a}b}{ab}.$
B. ${{\log }_{6}}45=\dfrac{2a-2\text{a}b}{ab}.$
C. ${{\log }_{6}}45=\dfrac{a+2\text{a}b}{ab+b}.$
D. ${{\log }_{6}}45=\dfrac{2{{\text{a}}^{2}}-2\text{a}b}{ab+b}.$
Ta có : ${{\log }_{6}}45={{\log }_{6}}9+{{\log }_{6}}5$
${{\log }_{6}}9=\dfrac{1}{{{\log }_{{{3}^{2}}}}\left( 2.3 \right)}=\dfrac{1}{\dfrac{1}{2}.\left( {{\log }_{3}}2+{{\log }_{3}}3 \right)}=\dfrac{2}{\dfrac{1}{{{\log }_{2}}3}+1}=\dfrac{2}{\dfrac{1}{a}+1}=\dfrac{2\text{a}}{a+1}\left( 1 \right)$ ${{\log }_{6}}5=\dfrac{1}{{{\log }_{5}}\left( 2.3 \right)}=\dfrac{1}{{{\log }_{5}}2+{{\log }_{5}}3}=\dfrac{1}{{{\log }_{5}}2+b}$ mà ${{\log }_{5}}2=\dfrac{{{\log }_{3}}2}{{{\log }_{3}}5}=\dfrac{\dfrac{1}{{{\log }_{2}}3}}{\dfrac{1}{{{\log }_{3}}5}}=\dfrac{\dfrac{1}{a}}{\dfrac{1}{b}}=\dfrac{b}{a}$
$\Rightarrow {{\log }_{6}}5=\dfrac{1}{\dfrac{b}{a}+b}=\dfrac{a}{ab+b}\left( 2 \right).$ Từ (1) và (2) suy ra : ${{\log }_{6}}45=\dfrac{2\text{a}}{a+1}+\dfrac{a}{ab+b}$
$=\dfrac{2{{\text{a}}^{2}}b+2\text{a}b+{{a}^{2}}+a}{\left( a+1 \right)\left( ab+b \right)}=\dfrac{\left( a+1 \right)2\text{a}b+\left( a+1 \right)a}{\left( a+1 \right)\left( ab+b \right)}=\dfrac{\left( a+1 \right)\left( a+2\text{a}b \right)}{\left( a+1 \right)\left( ab+b \right)}=\dfrac{a+2\text{a}b}{ab+b}.$
Cách khác:
Sử dụng lệnh gán Shift Sto. Gán $A={{\log }_{2}}3,B={{\log }_{5}}3$ bằng cách : Nhập ${{\log }_{2}}3$ bấm Shift Sto A tương tự B.
Thử từng đáp án: $\dfrac{A+2AB}{AB}-{{\log }_{6}}45\approx 1,34$ (Loại)
Thử đáp án: $\dfrac{A+2AB}{AB}-{{\log }_{6}}45=0$ (Đáp án).
${{\log }_{6}}9=\dfrac{1}{{{\log }_{{{3}^{2}}}}\left( 2.3 \right)}=\dfrac{1}{\dfrac{1}{2}.\left( {{\log }_{3}}2+{{\log }_{3}}3 \right)}=\dfrac{2}{\dfrac{1}{{{\log }_{2}}3}+1}=\dfrac{2}{\dfrac{1}{a}+1}=\dfrac{2\text{a}}{a+1}\left( 1 \right)$ ${{\log }_{6}}5=\dfrac{1}{{{\log }_{5}}\left( 2.3 \right)}=\dfrac{1}{{{\log }_{5}}2+{{\log }_{5}}3}=\dfrac{1}{{{\log }_{5}}2+b}$ mà ${{\log }_{5}}2=\dfrac{{{\log }_{3}}2}{{{\log }_{3}}5}=\dfrac{\dfrac{1}{{{\log }_{2}}3}}{\dfrac{1}{{{\log }_{3}}5}}=\dfrac{\dfrac{1}{a}}{\dfrac{1}{b}}=\dfrac{b}{a}$
$\Rightarrow {{\log }_{6}}5=\dfrac{1}{\dfrac{b}{a}+b}=\dfrac{a}{ab+b}\left( 2 \right).$ Từ (1) và (2) suy ra : ${{\log }_{6}}45=\dfrac{2\text{a}}{a+1}+\dfrac{a}{ab+b}$
$=\dfrac{2{{\text{a}}^{2}}b+2\text{a}b+{{a}^{2}}+a}{\left( a+1 \right)\left( ab+b \right)}=\dfrac{\left( a+1 \right)2\text{a}b+\left( a+1 \right)a}{\left( a+1 \right)\left( ab+b \right)}=\dfrac{\left( a+1 \right)\left( a+2\text{a}b \right)}{\left( a+1 \right)\left( ab+b \right)}=\dfrac{a+2\text{a}b}{ab+b}.$
Cách khác:
Sử dụng lệnh gán Shift Sto. Gán $A={{\log }_{2}}3,B={{\log }_{5}}3$ bằng cách : Nhập ${{\log }_{2}}3$ bấm Shift Sto A tương tự B.
Thử từng đáp án: $\dfrac{A+2AB}{AB}-{{\log }_{6}}45\approx 1,34$ (Loại)
Thử đáp án: $\dfrac{A+2AB}{AB}-{{\log }_{6}}45=0$ (Đáp án).
Đáp án C.