Cho a>0 , b>0 thỏa mãn log3a+2b+1(9a^2+b^2+1)+log6ab+1(3a+2b+10)=2 . Giá trị của a+2b bằng

Câu hỏi:

A. 6

B. 9

C. $\frac{7}{2}$

D. $\frac{5}{2}$

Trả lời:

Chọn C.

Ta có $a>0$ ,$b>0$  nên $\left\{\begin{array}{l}3a+2b+1>1\\ 9a^2+b^2+1>1\\ 6ab+1>1\end{array}\right.$$\Rightarrow \left\{\begin{array}{l}{\log }_{3a+2b+1}\left(9a^2+b^2+1\right)>0\\ {\log }_{6ab+1}\left(3a+2b+1\right)>0\end{array}\right.$

Áp dụng BĐT Cô-si cho hai số dương ta được

${\log }_{3a+2b+1}\left(9a^2+b^2+1\right)+{\log }_{6ab+1}\left(3a+2b+1\right)\ge 2\sqrt{{\log }_{3a+2b+1}\left(9a^2+b^2+1\right)+{\log }_{6ab+1}\left(3a+2b+1\right)}$

$\iff 2\ge 2\sqrt{{\log }_{6ab+1}\left(9a^2+b^2+1\right)}\iff {\log }_{6ab+1}\left(9a^2+b^2+1\right)\le 1\iff 9a^2+b^2+1\le 6ab+1$

$\iff {\left(3a-b\right)}^2\le 0\iff 3a=b$

Vì dấu “ ” đã xảy ra nên

${\log }_{3a+2b+1}\left(9a^2+b^2+1\right)={\log }_{6ab+1}\left(3a+2b+1\right)\iff {\log }_{3b+1}\left(2b^2+1\right)={\log }_{2b^2+1}\left(3b+1\right)$

 $\iff 2b^2+1=3b+1\iff b=\frac{3}{2}$(vì $b>0$ ). Suy ra $a=\frac{1}{2}$  .

Vậy $a+2b=\frac{1}{2}+3$$=\frac{7}{2}$ .