Phương trình log 4 (x +1) ^2 + 2 = log căn bậc hai 2 căn bậc hai 4 - x + log 4 (4 + x) ^3

* Hướng dẫn giải

Chọn C

Điều kiện: $\left\{\begin{array}{l}x+1\ne 0\\ 4-x>0\\ 4+x>0\end{array}\right.\iff x\in \left(-4;4\right)\backslash \left\{-1\right\}$.

Khi đó, $PT\iff {\log }_{2^2}{\left(x+1\right)}^2+2={\log }_{2^{\frac{1}{2}}}{\left(4-x\right)}^{\frac{1}{2}}+{\log }_{2^2}{\left(4+x\right)}^3$

$$\begin{gathered} \iff {\log }_2\left|x+1\right|+{\log }_{2}4={\log }_2\left(4-x\right)+{\log }_2\left(x+4\right) \\ \iff {\log }_{2}4\left|x+1\right|={\log }_2\left(16-x^2\right)\iff 4\left|x+1\right|=16-x^2\left(*\right) \end{gathered}$$

* TH1: $x+1>0\Rightarrow -1<x<4$: Ta có $\left(*\right)\iff 4x+4=16-x^2\iff x^2+4x-12=0$

$\iff \left[\begin{array}{l}x=2\left(tm\right)\\ x=-6\left(l\right)\end{array}\right.\Rightarrow x_1=2$.

* TH2: $x+1<0\Rightarrow -4<x<-1$: $\left(*\right)\iff -4x-4=16-x^2\iff x^2-4x-20=0$

$\iff \left[\begin{array}{l}x=2+2\sqrt{6}\left(l\right)\\ x=2-2\sqrt{6}\left(tm\right)\end{array}\right.\Rightarrow x_2=2-2\sqrt{6}$.

Vậy $\left|x_1-x_2\right|=2\sqrt{6}$.