Ta có : \(\left|x+2,1\right|-\left| x-2,1\right|=1,2\) (1)
Xét :
1. Nếu \(x>2,1\Rightarrow\left(1\right)\Leftrightarrow x+2,1-\left(x-2,1\right)=1,2\)
\(\Leftrightarrow4,2=1,2\left(\text{vô lí - loại}\right)\)
2. Nếu \(x< -2,1\) \(\Rightarrow\left(1\right)\Leftrightarrow-\left(x+2,1\right)-\left(2,1-x\right)=1,2\)
\(\Leftrightarrow-4,2=1,2\left(\text{vô lí - loại}\right)\)
3. Nếu \(-2,1\le x\le2,1\) \(\Rightarrow\left(1\right)\Leftrightarrow\left(x+2,1\right)-\left(2,1-x\right)=1,2\)
\(\Leftrightarrow2x=1,2\Leftrightarrow x=0,6\left(\text{thỏa mãn}\right)\)
Vậy x = 0,6
\(\left|x+2,1\right|-\left|x-2,1\right|=1,2\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2,1=1,2\\x-2,1=1,2\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-0,9\\x=3,3\end{array}\right.\)
Vậy \(x\in\left\{-0,9;3,3\right\}\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x+2,1\right|-\left|x-2,1\right|\ge\left|x+2,1-x-2,1\right|=0< VP\)
=>ko tồn tại x