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Câu hỏi: Gọi $x, y$ là các số thực dương thỏa mãn điều kiện $\log _9 x=\log _6 y=\log _4(x+y)$ và $\dfrac{x}{y}=\dfrac{-a+\sqrt{b}}{2}$, với $a, b$ là hai số nguyên dương. Tính $T=a+b$.
A. $T=6$.
B. $T=8$.
C. $T=11$.
D. $T=4$.
A. $T=6$.
B. $T=8$.
C. $T=11$.
D. $T=4$.
Đặt $t=\log _9 x=\log _6 y=\log _4(x+y) \rightarrow\left\{\begin{array}{l}x=9^t \\ y=6^t \\ x+y=4^t\end{array}\right.$.
Suy ra $x+y=4^t \Leftrightarrow 9^t+6^t=4^t \Leftrightarrow\left(\dfrac{3}{2}\right)^{2 t}+\left(\dfrac{3}{2}\right)^t=1 \Leftrightarrow\left(\dfrac{3}{2}\right)^{2 t}+\left(\dfrac{3}{2}\right)^t-1=0$.
$\Leftrightarrow\left(\dfrac{3}{2}\right)^t=\dfrac{-1+\sqrt{5}}{2}$.
Mặt khác $\dfrac{x}{y}=\dfrac{9^t}{6^t}=\left(\dfrac{9}{6}\right)^t=\left(\dfrac{3}{2}\right)^t=\dfrac{-a+\sqrt{b}}{2}$ suy ra $\dfrac{-a+\sqrt{b}}{2}=\dfrac{-1+\sqrt{5}}{2} \rightarrow\left\{\begin{array}{l}a=1 \\ b=5\end{array}\right.$.
Vậy $T=a+b=6$.
Suy ra $x+y=4^t \Leftrightarrow 9^t+6^t=4^t \Leftrightarrow\left(\dfrac{3}{2}\right)^{2 t}+\left(\dfrac{3}{2}\right)^t=1 \Leftrightarrow\left(\dfrac{3}{2}\right)^{2 t}+\left(\dfrac{3}{2}\right)^t-1=0$.
$\Leftrightarrow\left(\dfrac{3}{2}\right)^t=\dfrac{-1+\sqrt{5}}{2}$.
Mặt khác $\dfrac{x}{y}=\dfrac{9^t}{6^t}=\left(\dfrac{9}{6}\right)^t=\left(\dfrac{3}{2}\right)^t=\dfrac{-a+\sqrt{b}}{2}$ suy ra $\dfrac{-a+\sqrt{b}}{2}=\dfrac{-1+\sqrt{5}}{2} \rightarrow\left\{\begin{array}{l}a=1 \\ b=5\end{array}\right.$.
Vậy $T=a+b=6$.
Đáp án A.