\(x^2\ge0\Rightarrow x^2+1\ge1>0\Rightarrow\left|x^2+1\right|=x^2+1\)
<=>\(x^2+1-\left|x^2-4\right|=1\Leftrightarrow x^2-\left|x^2-4\right|=0\Leftrightarrow x^2=\left|x^2-4\right|\)
+)\(x^2-4>0\Leftrightarrow x^2>4\Leftrightarrow x< -2;x>2\)
<=>\(x^2-4=x^2\Leftrightarrow0=4\) vô lý
+)\(x^2-4\le0\Leftrightarrow x^2\le4\Leftrightarrow-2\le x\le2\)
<=>\(4-x^2=x^2\Leftrightarrow4=2x^2\Leftrightarrow x^2=2\Leftrightarrow\orbr{\begin{cases}x=-\sqrt{2}\\x=\sqrt{2}\end{cases}}\)(nhận)
Vậy ...