Cho biểu thức

D

- Hướng dẫn giải

Đáp án D

Ta có

$\begin{array}{l}A=\log \left(2017+\log \left(2016+\log \left(2015+\log \left(\mathrm{...}+\log \left(3+\log 2\right)\mathrm{...}\right)\right)\right)\right)\\ >\log \left(2017+\log 2016\right)>\log \left(2017+3\right)=\log 2010\Rightarrow A>\log 2010\end{array}$

Áp dụng bất đẳng thức $\log x<x,\forall x>1,$ ta có

$\begin{array}{l}2015+\log \left(2014+\log \left(\mathrm{...}+\log \left(3+\log 2\right)\mathrm{...}\right)\right)<2015+2014+\log \left(\mathrm{...}+\log \left(3+\log 2\right)\mathrm{...}\right)\\ \text{ < 2015+1014+2013+}\mathrm{...}\text{+3+2=}\frac{2017\times 2014}{2}\end{array}$

Khi đó 

$\begin{array}{l}\log \left(2016+\log \left(2015+\log \left(2014+\log \left(\mathrm{...}+\log \left(3+\log 2\right)\mathrm{...}\right)\right)\right)\right)\\ <\log \left(2016+\frac{2017\times 2014}{2}\right)<4\end{array}$

Vậy $A<\log \left(2017+4\right)=\log 2021\stackrel{}{\to }A\in \left(\log 2010;2021\right)$