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Câu hỏi: Cho a, b > 0 thỏa mãn ${{\log }_{2a+3b+1}}(25{{a}^{2}}+{{b}^{2}}+1)+{{\log }_{10ab+1}}(2a+3b+1)=2$. Giá trị của $a+4b$ bằng
A. $5$
B. $6$
C. $\dfrac{357}{50}$
D. $\dfrac{407}{50}$
A. $5$
B. $6$
C. $\dfrac{357}{50}$
D. $\dfrac{407}{50}$
Với $a,b>0\Rightarrow \left\{ \begin{aligned}
& 25{{a}^{2}}+{{b}^{2}}+1>1 \\
& 2a+3b+1>1 \\
& 10ab+1>0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\log }_{2a+3b+1}}(25{{a}^{2}}+{{b}^{2}}+1)>0 \\
& {{\log }_{10ab+1}}(2a+3b+1)>0 \\
\end{aligned} \right.$
Ta có $P={{\log }_{2a+3b+1}}(25{{a}^{2}}+{{b}^{2}}+1)+{{\log }_{10ab+1}}(2a+3b+1)$
$\ge {{\log }_{2a+3b+1}}(10ab+1)+lo{{g}_{10ab+1}}(2a+3b+1)$
$\ge 2\sqrt{{{\log }_{2a+3b+1}}(10ab+1).lo{{g}_{10ab+1}}(2a+3b+1)}=2$
Dấu "=" xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& 5a=b \\
& {{\log }_{2a+3b+1}}(10ab+1)=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 5a=b \\
& 10ab+1=2a+3b+1 \\
\end{aligned} \right.$
$\Rightarrow 50{{a}^{2}}=2a+15a\Rightarrow a=\dfrac{17}{50}\Rightarrow b=\dfrac{17}{10}$
& 25{{a}^{2}}+{{b}^{2}}+1>1 \\
& 2a+3b+1>1 \\
& 10ab+1>0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\log }_{2a+3b+1}}(25{{a}^{2}}+{{b}^{2}}+1)>0 \\
& {{\log }_{10ab+1}}(2a+3b+1)>0 \\
\end{aligned} \right.$
Ta có $P={{\log }_{2a+3b+1}}(25{{a}^{2}}+{{b}^{2}}+1)+{{\log }_{10ab+1}}(2a+3b+1)$
$\ge {{\log }_{2a+3b+1}}(10ab+1)+lo{{g}_{10ab+1}}(2a+3b+1)$
$\ge 2\sqrt{{{\log }_{2a+3b+1}}(10ab+1).lo{{g}_{10ab+1}}(2a+3b+1)}=2$
Dấu "=" xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& 5a=b \\
& {{\log }_{2a+3b+1}}(10ab+1)=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 5a=b \\
& 10ab+1=2a+3b+1 \\
\end{aligned} \right.$
$\Rightarrow 50{{a}^{2}}=2a+15a\Rightarrow a=\dfrac{17}{50}\Rightarrow b=\dfrac{17}{10}$
Đáp án C.