12. \(\Delta=4\left(m-1\right)^2-4\left(m^2-2m-8\right)=0=36\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{2m-8}{2}=m-4\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{2m+4}{2}=m+2\end{matrix}\right.\)
Ta có: \(\sqrt{x_1}=x_2+4\)
\(TH_1:x_1=m-4\left(m\ge4\right)\Rightarrow x_2=m+2\)
\(\Rightarrow\sqrt{m-4}=m+2+4\Rightarrow m+6-\sqrt{m-4}=0\)
\(\Leftrightarrow m-4-\sqrt{m-4}+10=0\)
Đặt \(t=\sqrt{m-4}\)
pt trở thành \(t^2-t+10=0\)
mà \(t^2-t+10=t^2-t+\dfrac{1}{4}+\dfrac{39}{10}=\left(t-\dfrac{1}{2}\right)^2+\dfrac{39}{10}>0\)
\(\Rightarrow\) loại
\(TH_2:x_2=m-4\Rightarrow x_1=m+2\left(m\ge-2\right)\)
\(\Rightarrow\sqrt{m+2}=m\Rightarrow m+2=m^2\Rightarrow m^2-m-2=0\)
\(\Leftrightarrow\left(m-2\right)\left(m+1\right)=0\Rightarrow\left[{}\begin{matrix}m=2\\m=-1\end{matrix}\right.\)
