Có bao nhiêu số nguyên x  thỏa mãn

$\left[{\log }_3\left(x^2+1\right)-{\log }_3\left(x+31\right)\right]\left(32-2^{x-1}\right)\ge 0$?

Đáp án đúng là: A

$\left[{\log }_3\left(x^2+1\right)-{\log }_3\left(x+31\right)\right]\left(32-2^{x-1}\right)\ge 0\iff \left[\begin{array}{l}\left\{\begin{array}{l}{\log }_3\left(x^2+1\right)-{\log }_3\left(x+31\right)\ge 0\\ 32-2^{x-1}\ge 0\end{array}\right.\\ \left\{\begin{array}{l}{\log }_3\left(x^2+1\right)-{\log }_3\left(x+31\right)\le 0\\ 32-2^{x-1}\le 0\end{array}\right.\end{array}\right.$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}{\log }_3\left(x^2+1\right)\ge {\log }_3\left(x+31\right)\\ 2^{x-1}\le 32\end{array}\right.\\ \left\{\begin{array}{l}{\log }_3\left(x^2+1\right)\le {\log }_3\left(x+31\right)\\ 2^{x-1}\ge 32\end{array}\right.\end{array}\right.$$\iff \left[\begin{array}{l}\left\{\begin{array}{l}{x}^2+1\ge x+31\\ x+31>0\\ 2^{x-1}\le 2^5\end{array}\right.\\ \left\{\begin{array}{l}{x}^2+1\le x+31\\ 2^{x-1}\ge 2^5\\ x^2+1\ge 0\text{\hspace{0.17em}\hspace{0.17em}(luôn\hspace{0.17em}\hspace{0.17em}đúng)}\end{array}\right.\end{array}\right.$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}{x}^2-x-30\ge 0\\ x>-31\\ x\le 6\end{array}\right.\\ \left\{\begin{array}{l}{x}^2-x-30\le 0\\ x\ge 6\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}-31<x\le -5\\ x=6\end{array}\right.$

Mà $x\in \mathbb{Z}$  nên $x\in \left\{-30;-29;-28;\mathrm{...};-6;-5;6\right\}$

Vậy có 27  số nguyên x  thỏa mãn đề bài.