m và n là các số tự nhiên nguyên tố cùng nhau. Tìm tổng m + n

* Hướng dẫn giải

Đáp án C

*  ${\log }_{\frac{1}{16}}x$ xác định khi $x>0$

*  ${\log }_{16}\left({\log }_{\frac{1}{16}}x\right)$ xác định khi  ${\log }_{\frac{1}{16}}x>0={\log }_{\frac{1}{16}}1$$\iff 0<x<1$

*  ${\log }_{\frac{1}{4}}\left({\log }_{16}\left({\log }_{\frac{1}{16}}x\right)\right)$ xác định khi

$\begin{array}{l}{\log }_{16}\left({\log }_{\frac{1}{16}}x\right)>0={\log }_{16}1\\ \Rightarrow {\log }_{\frac{1}{16}}x>1={\log }_{\frac{1}{16}}\frac{1}{16}\\ \Rightarrow x<\frac{1}{16}\end{array}$

*  ${\log }_4\left({\log }_{\frac{1}{4}}\left({\log }_{16}\left({\log }_{\frac{1}{16}}x\right)\right)\right)$ xác định khi

$\begin{array}{l}{\log }_{\frac{1}{4}}\left({\log }_{16}\left({\log }_{\frac{1}{16}}x\right)\right)>0={\log }_{\frac{1}{4}}1\\ \Rightarrow {\log }_{16}\left({\log }_{\frac{1}{16}}x\right)<1={\log }_{16}16\end{array}$ 

$\begin{array}{l}\Rightarrow {\log }_{\frac{1}{16}}x<16={\log }_{\frac{1}{16}}\frac{1}{{16}^{16}}\\ \Rightarrow x>\frac{1}{{16}^{16}}\end{array}$

*  ${\log }_{\frac{1}{2}}\left({\log }_4\left({\log }_{\frac{1}{4}}\left({\log }_{16}\left({\log }_{\frac{1}{16}}x\right)\right)\right)\right)$xác định khi

${\log }_4\left({\log }_{\frac{1}{4}}\left({\log }_{16}\left({\log }_{\frac{1}{16}}x\right)\right)\right)>0={\log }_{4}1$

$\begin{array}{l}\Rightarrow {\log }_{\frac{1}{4}}\left({\log }_{16}\left({\log }_{\frac{1}{16}}x\right)\right)>1={\log }_{\frac{1}{4}}\frac{1}{4}\\ \Rightarrow {\log }_{16}\left({\log }_{\frac{1}{16}}x\right)<\frac{1}{4}={\log }_{16}2\end{array}$

 $\begin{array}{l}\Rightarrow {\log }_{\frac{1}{16}}x<2={\log }_{\frac{1}{16}}\frac{1}{{16}^2}\\ \Rightarrow x>\frac{1}{{16}^2}\end{array}$

Kết hợp tất cả các điều kiện ta được

$\begin{array}{l}\frac{1}{{16}^2}<x<\frac{1}{16}\\ \Rightarrow D=\left(\frac{1}{{16}^2};\frac{1}{16}\right)\\ \Rightarrow b-a=\frac{15}{256}\\ \Rightarrow m+n=271\end{array}$