1-(x-2)-[(x-3)-(x-1)]=0

Given, $\quad 1-(x-2)-[(x-3)-(x-1)]=0$

$\Rightarrow$ $1-x+2-[x-3-x+1]=0$

$\Rightarrow \quad 3-x-[-2]=0$

$\Rightarrow \quad 3-x+2=0$

$\Rightarrow \quad-x+5=0$

$\Rightarrow$ $-x+5=0$

$\Rightarrow$ $-x=-5$  [transposing 5 to RHS]

$\Rightarrow$ $\frac{-x}{-1}=\frac{-5}{-1}$ [dividing both sides by -1]

$\therefore$ $x=5$