Ta có : \(x-\left\{x-\left[x+x-1\right]\right\}=1\)
\(x-\left\{x-x-x+1\right\}=1\)
\(x-x+x+x-1=1\)
\(2x-1=1\)
\(2x=1+1\)
\(2x=2\)
\(\Rightarrow x=2:2\)
\(x=1\)
\(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)
\(x-\left\{x-\left[x+x-1\right]\right\}=1\)
\(x-\left\{x-x-x+1\right\}=1\)
\(x-x+x+x-1=1\)
\(2x-1=1\)
\(2x=2\)
\(x=1\)
Vậy x=1