smc

We have $\sqrt{x^2}=|x|$, so we rewrite the expression as follows. $$|x-\sqrt{(x-1{)}^2}|=|x-|x-1||$$ We know that $x<0$, so $x-1<0$. Thus, we can rewrite $|x-1|$ as $1-x$. So $$|x-|x-1||=|x-(1-x)|=|2x-1|$$. Since $x<0,2x-1<0$. Thus, we can write this as $$|2x-1|=1-2x$$ $\boxed{1-2x}$