Phương trình log 4 (x+1)^2+2=log căn2 căn(4-x)+log 8 (4+x)^3

* Hướng dẫn giải

Đáp án C

Điều kiện $\left\{\begin{array}{l}{\left(x+1\right)}^2>0\\ 4-x>0\\ {\left(4+x\right)}^3>0\end{array}\right.\iff \left\{\begin{array}{l}x\ne -1\\ x<4\\ x>-4\end{array}\right.\iff \left\{\begin{array}{l}-4<x<4\\ x\ne -1\end{array}\right.$ 

$$\begin{gathered} PT\iff {\log }_2\sqrt{{\left(x+1\right)}^2}+2={\log }_2\left(4-x\right)+{\log }_2\left(4+x\right) \\ \iff {\log }_2\left[4\sqrt{{\left(x+1\right)}^2}\right]={\log }_2\left(4-x\right)\left(4+x\right) \end{gathered}$$

 $\begin{array}{l}\iff 4\sqrt{{\left(x+1\right)}^2}=16-x^2\iff \left[\begin{array}{l}\left\{\begin{array}{l}x+1\ge 0\\ 4\left(x+1\right)=16-x^2\end{array}\right.\\ \left\{\begin{array}{l}x+1<0\\ -4\left(x+1\right)=16-x^2\\ \end{array}\right.\end{array}\right.\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}x\ge -1\\ x^2+4x-12=0\end{array}\right.\\ \left\{\begin{array}{l}x<-1\\ x^2-4x-20=0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x\ge -1\\ \left[\begin{array}{l}x=2\\ x=-6\end{array}\right.\end{array}\right.\\ \left\{\begin{array}{l}x<-1\\ \left[\begin{array}{l}x=2+2\sqrt{6}\\ x=2-2\sqrt{6}\end{array}\right.\end{array}\right.\end{array}\right.\\ \iff \left[\begin{array}{l}x=2\\ x=2-2\sqrt{6}\end{array}\right.\end{array}$