Cho \(a>0,\ b>0\) thỏa mãn \({{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{4ab+1}}\left( 2a+2b+1 \right)=2.\) Giá trị của \(a+2b\) bằng:

A. \(\frac{15}{4}\)

B. \(5\)

C. \(4\)

D. \(\frac{3}{2}\)

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Ta có: \({{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{4ab+1}}\left( 2a+2b+1 \right)=2\)

\(\Leftrightarrow {{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)+\frac{1}{{{\log }_{2a+2b+1}}\left( 4ab+1 \right)}=2.\)

Có: \({{\left( 2a \right)}^{2}}+{{b}^{2}}\ge 2.2a.b\Leftrightarrow 4{{a}^{2}}+{{b}^{2}}\ge 4ab.\)

\(\Rightarrow 4{{a}^{2}}+{{b}^{2}}+1\ge 4ab+1.\)

Dấu “=” xảy ra \(\Leftrightarrow 2a=b.\)

Theo giả thiết ta có:

\(\left\{ \begin{array}{l}a > 0\\b > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}2a + 2b + 1 > 1\\4a + 1 > 1\end{array} \right.\) \(\Rightarrow \left\{ \begin{array}{ccccc}\log _{2a + 2b + 1}\left( {4{a^2} + {b^2} + 1} \right) \ge {\log _{2a + 2b + 1}}\left( {4ab + 1} \right)\\{\log _{2a + 2b + 1}}\left( {2a + 2b + 1} \right) = \frac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}}\end{array} \right..\)

Áp dụng bất đẳng thức Cauchy ta có:

\(\begin{align} & {{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)+\frac{1}{{{\log }_{2a+2b+1}}\left( 4ab+1 \right)}\le {{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)+\frac{1}{{{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)} \\ & \le 2.\sqrt{{{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right).\frac{1}{{{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)}}=2. \\\end{align}\)

Dấu “=” xảy ra \(\Leftrightarrow \left\{ \begin{align} & 2a=b \\ & {{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)=\frac{1}{{{\log }_{2a+2b+1}}\left( 4{{a}^{2}}+{{b}^{2}}+1 \right)} \\\end{align} \right.\)

\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
2a = b\\
{\log _{2a + 2b + 1}}\left( {4{a^2} + {b^2} + 1} \right) = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2a = b\\
{\log _{3b + 1}}\left( {2{b^2} + 1} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 2a\\
2{b^2} + 1 = 3b + 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 2a\\
2{b^2} - 3b = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 2a\\
\left[ \begin{array}{l}
b = 0\;\;\left( {ktm} \right)\\
b = \frac{3}{2}\;\;\left( {tm} \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = \frac{3}{4}\;\;\left( {tm} \right)\\
b = \frac{3}{2}
\end{array} \right..
\end{array}\)