Cho $a>0, b>0$ thỏa mãn $\log _4 a=\log _{25} b=\log \dfrac{4...

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\begin{aligned}
& \text { Ta có } \log _4 a=\log _{25} b=\log \dfrac{4 b-a}{4}=t \Rightarrow\left\{\begin{array} { l }
{ a = 4 ^ { t } } \\
{ b = 2 5 ^ { t } } \\
{ \dfrac { 4 b - a } { 4 } = 1 0 ^ { t } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=4^t(1) \\
b=25^t(2) \\
\dfrac{4.25^t-4^t}{4}=10^t(3)
\end{array}\right.\right. \\
& \text { Có (3) } \Leftrightarrow 4.25^t-4^t=4.10^t \Leftrightarrow\left(2^t\right)^2+4.2^t .5^t-4 .\left(5^t\right)^2=0 \\
& \Leftrightarrow\left[\left(\dfrac{2}{5}\right)^t\right]^2+4\left(\dfrac{2}{5}\right)^t-4=0 \Leftrightarrow\left[\begin{array}{l}
\left(\dfrac{2}{5}\right)^t=-2+2 \sqrt{2} \\
\left(\dfrac{5}{2}\right)^t=-2-2 \sqrt{2}(V N)
\end{array} \Rightarrow \dfrac{a}{b}=\left(\dfrac{2}{5}\right)^{2 t}=(2 \sqrt{2}-2)^2\right. \\
& =12-8 \sqrt{2} \\
&
\end{aligned}
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\begin{gathered}
\Rightarrow \log _{\sqrt{6}}\left(\dfrac{a}{2}+4 b \sqrt{2}\right)-\log _{\sqrt{6}} b=\log _{\sqrt{6}}\left(\dfrac{\dfrac{a}{2}+4 b \sqrt{2}}{b}\right)=\log _{\sqrt{6}}\left(\dfrac{a}{2 b}+4 \sqrt{2}\right) \\
=\log _{\sqrt{6}}(6-4 \sqrt{2}+4 \sqrt{2})=\log _{\sqrt{6}} 6=2
\end{gathered}
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