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Câu hỏi: Cho a, b là các số dương thỏa mãn ${{\log }_{9}}a={{\log }_{16}}b={{\log }_{12}}\dfrac{5b-a}{2}$. Tính giá trị $\dfrac{a}{b}$.
A. $\dfrac{a}{b}=\dfrac{3+\sqrt{6}}{4}$
B. $\dfrac{a}{b}=7-2\sqrt{6}$
C. $\dfrac{a}{b}=7+2\sqrt{6}$
D. $\dfrac{a}{b}=\dfrac{3-\sqrt{6}}{4}$
A. $\dfrac{a}{b}=\dfrac{3+\sqrt{6}}{4}$
B. $\dfrac{a}{b}=7-2\sqrt{6}$
C. $\dfrac{a}{b}=7+2\sqrt{6}$
D. $\dfrac{a}{b}=\dfrac{3-\sqrt{6}}{4}$
Đặt ${{\log }_{9}}a={{\log }_{16}}b={{\log }_{12}}\dfrac{5b-a}{2}=t$, ta được: $a={{9}^{t}},b={{16}^{t}}=\dfrac{5b-a}{2}={{12}^{t}}$
$\Rightarrow \dfrac{{{5.16}^{t}}-{{9}^{t}}}{2}={{12}^{t}}\Leftrightarrow {{5.16}^{t}}-{{2.12}^{t}}-{{9}^{t}}=0\Leftrightarrow 5-2.{{\left( \dfrac{3}{4} \right)}^{t}}-{{\left( \dfrac{3}{4} \right)}^{2t}}=0\Leftrightarrow {{\left( \dfrac{3}{4} \right)}^{t}}=\sqrt{6}-1$.
Do đó $\dfrac{a}{b}=\dfrac{{{9}^{t}}}{{{16}^{t}}}={{\left( \dfrac{3}{4} \right)}^{2t}}={{\left( \sqrt{6}-1 \right)}^{2}}=7-2\sqrt{6}$.
$\Rightarrow \dfrac{{{5.16}^{t}}-{{9}^{t}}}{2}={{12}^{t}}\Leftrightarrow {{5.16}^{t}}-{{2.12}^{t}}-{{9}^{t}}=0\Leftrightarrow 5-2.{{\left( \dfrac{3}{4} \right)}^{t}}-{{\left( \dfrac{3}{4} \right)}^{2t}}=0\Leftrightarrow {{\left( \dfrac{3}{4} \right)}^{t}}=\sqrt{6}-1$.
Do đó $\dfrac{a}{b}=\dfrac{{{9}^{t}}}{{{16}^{t}}}={{\left( \dfrac{3}{4} \right)}^{2t}}={{\left( \sqrt{6}-1 \right)}^{2}}=7-2\sqrt{6}$.
Đáp án B.