B1. Simplify the expression: $\frac{2-x}{x^{3}-x^{2}-x+1}:\left(\frac{1}{x-1} \cdot \frac{x}{x+1}-\frac{2}{x+1}\right)$.
B1. First, we factor the denominator of the first fraction $x^{3}-x^{2}-x+1=(x-1)^{2}(x+1)$, then simplify the expression in the parentheses: $\frac{-x+2}{(x+1)(x-1)}$. Finally, we perform the division of fractions as multiplication by the reciprocal $\frac{-x+2}{(x-1)^{2}(x+1)} \cdot \frac{(x+1)(x-1)}{-x+2}$, and after canceling out, we get $\frac{1}{x-1}$.

Factorization of the first denominator $x^{3}-x^{2}-x+1=(x-1)^{2}(x+1) \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . . .1$ point

Determination of the common denominator of the expression in the parentheses ..................................................................................................

Expansion to the common denominator ................................................................................$\qquad$
Result $\frac{1}{x-1}$

1 point