Cho các số thực dương a, b thỏa mãn log4a=log6b=log9(4a−5b)−1. Đặt T=b/a

Hướng dẫn giải:

Đáp án A.

Đặt  \( t={{\log }_{9}}x={{\log }_{6}}y={{\log }_{4}}(x+y) \) , ta có: \( \left\{ \begin{align}& x={{9}^{t}} \\ & y={{6}^{t}} \\& x+y={{4}^{t}} \\\end{align} \right.\Rightarrow {{9}^{t}}+{{6}^{t}}={{4}^{t}} \)

 \( \Leftrightarrow {{\left( \frac{3}{2} \right)}^{2t}}+{{\left( \frac{3}{2} \right)}^{t}}-1=0 \) \( \Leftrightarrow \left\{ \begin{align}& {{\left( \frac{3}{2} \right)}^{t}}=\frac{-1-\sqrt{5}}{2}\text{(loại)} \\& {{\left( \frac{3}{2} \right)}^{t}}=\frac{-1+\sqrt{5}}{2} \\\end{align} \right. \)

 \( \Rightarrow {{\left( \frac{3}{2} \right)}^{t}}=\frac{-1+\sqrt{5}}{2} \)

Suy ra:  \( \frac{x}{y}={{\left( \frac{9}{6} \right)}^{t}}={{\left( \frac{3}{2} \right)}^{t}}=\frac{-1+\sqrt{5}}{2}=\frac{-a+\sqrt{b}}{2} \)

 \( \Rightarrow \left\{ \begin{align}& a=1 \\& b=5 \\\end{align} \right.\Rightarrow T={{a}^{2}}+{{b}^{2}}={{1}^{2}}+{{5}^{2}}=26 \)