Ta có: \(f\left(x\right)=-2x+1\)
\(\Rightarrow\hept{\begin{cases}f\left(1-2x\right)=-2\left(1-2x\right)+1=4x-1\\f\left(2-x\right)=-2\left(2-x\right)+1=2x-3\end{cases}}\)
Vì \(f\left(1-2x\right)=f\left(2-x\right)\Rightarrow f\left(1-2x\right)-f\left(2-x\right)=0\)
Để \(f\left(1-2x\right)-f\left(2-x\right)=0\Rightarrow4x-1-\left(2x-3\right)=0\)
\(\Rightarrow4x-1-2x+3=0\Rightarrow2x+2=0\Rightarrow2\left(x+1\right)=0\)
\(\Rightarrow x+1=\frac{0}{2}=0\Rightarrow x=0-1\Rightarrow x=-1\)