ĐK $x>-1$
$\log_2^2 (x+1) -6\log_2 (x+1)^{\frac{1}{2}}+2=0$
$\Leftrightarrow \log_2^2 (x+1) -3\log_2 (x+1)+2=0$ đặt $\log_2 (x+1)=t$
$\Rightarrow t^2 -3t +2 =0 \Leftrightarrow t = 1;\ t = 2$
+ $t = 1 = \log_2 (x+1) \Rightarrow x+ 1 = 2 \Rightarrow x= 1$
+ $t=2= \log_2 (x+1) \Rightarrow x+ 1 = 4 \Rightarrow x= 3$