giải
\(\left|x^2-4\right|=x-3\Rightarrow\orbr{\begin{cases}x^2-4=-x+3\\x^2-4=x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+x=7\\x^2-x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\left(x+1\right)=7\\x\left(x-1\right)=1\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=8\\\left(x-1\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x\in\left\{\sqrt{8}-1;-\sqrt{8}-1\right\}\\x=1\end{cases}}}\)