Cho $\Large a > 0, b > 0$ thỏa mãn $\Large \mathrm{log}_{4a+5b+1}(16a^

Cho $\Large a > 0, b > 0$ thỏa mãn $\Large \mathrm{log}_{4a+5b+1}(16a^2+b^2+1)+\mathrm{log}_{8ab+1}(4a+5b+1)=2.$ Giá trị của $\Large a+2b$ bằng

• A.

  $\Large \dfrac{27}{4}.$

• B.

  6.

• C.

  $\Large \dfrac{20}{3}.$

• D.

  9.

Chọn A

Ta có: $\Large a > 0, b > 0$

Nên $\Large \left\{\begin{align} & 4a+5b+1 > 1 \\ & 8ab+1 > 1 \end{align}\right.$ $\Large \Rightarrow \left\{\begin{align} & \mathrm{log}_{4a+5b+1}(16a^2+b^2+1) > 0 \\ & \mathrm{log}_{8ab+1}(4a+5b+1) > 0 \end{align}\right.$

$\Large P=\mathrm{log}_{4a+5b+1}(16a^2+b^2+1)$+$\Large \mathrm{log}_{8ab+1}(4a+5b+1)$ $\Large \geq 2\sqrt{\mathrm{log}_{4a+5b+1}(16a^2+b^2+1).\mathrm{log}_{8ab+1}(4a+5b+1)}$

$\Large \Leftrightarrow P \geq 2\sqrt{\mathrm{log}_{8ab+1}(16a^2+b^2+1)}$

Mặt khác:

$\Large 16a^2+b^2+1 \geq 2\sqrt{16a^2b^2}+1=8ab+1$ $\Large \Leftrightarrow P \geq 2\sqrt{\mathrm{log}_{8ab+1}(8ab+1)}=2$

Dấu bằng xảy ra khi và chỉ khi: $\Large \left\{\begin{align} & 16a^2=b^2 \\ & 8ab+1=4a+5b+1 \end{align}\right.$ $\Large \Leftrightarrow \left\{\begin{align} & 4a=b \\ & 2b^2+1=6b+1 \end{align}\right.$ $\Large \Leftrightarrow \left\{\begin{align} & a=\dfrac{3}{4} \\ & b=3 \end{align}\right.$

Do đó $\Large a+2b=\dfrac{27}{4}.$