+ x2 < 1 ta có: 1 - x2 + |x| = 1
=> |x| - x2 = 0
=> |x| - |x|.|x| = 0
=> |x|.(1 - |x|) = 0
\(\Rightarrow\left[\begin{matrix}\left|x\right|=0\\1-\left|x\right|=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
+ x2 \(\ge\) 1, ta có: x2 - 1 + |x| = 1
=> x2 + |x| - 2 = 0
\(\Rightarrow\left[\begin{matrix}x^2+x-2=0\\x^2-x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\\\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x+\frac{1}{2}=\frac{3}{2}\\x+\frac{1}{2}=-\frac{3}{2}\\x-\frac{1}{2}=\frac{3}{2}\\x-\frac{1}{2}=-\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=-2\\x=2\\x=-1\end{matrix}\right.\) . Ta có: x = 1 và x = -1 thỏa mãn
Vậy ...
