Cho hàm số y= căn bậc hai log 1/5

* Hướng dẫn giải

Điều kiện ${\log }_{\frac{1}{5}}\left({\log }_5\frac{x^2+1}{x+3}\right)\ge 0={\log }_{\frac{1}{5}}1\iff 0<{\log }_5\frac{x^2+1}{x+3}\le 1$

$\iff {\log }_{5}1<{\log }_5\frac{x^2+1}{x+3}\le {\log }_{5}5\iff 1<\frac{x^2+1}{x+3}\le 5$

$$\begin{gathered} \iff \left\{\begin{array}{l}\frac{x^2-x-2}{x+3}>0\\ \frac{x^2-5x-14}{x+3}\le 0\end{array}\right.\iff \left\{\begin{array}{l}\left[\begin{array}{l}-3<x<-1\\ x>2\end{array}\right.\\ \left[\begin{array}{l}x<-3\\ -2\le x\le 7\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}-2\le x<-1\\ 2<x\le 7\end{array}\right. \\ \Rightarrow D=\left[-2;-1\right)\cup \left(2;7\right]. \end{gathered}$$

Khi đó:$\left\{\begin{array}{l}x\in D\\ x\in \mathbb{Z}\end{array}\right.\to x\in \left\{-2;3;4;5;6;7\right\}:$ có 6 số nguyên

Chọn B