Докажите тождество: (b+2)/(b^2-2b+1)∶(b^2-4)/(3b-3)-3/(b-2)=3/(1-b).

Ответ:

\[\frac{b + 2}{b^{2} - 2b + 1}\ :\frac{b^{2} - 4}{3b - 3} - \frac{3}{b - 2} =\]

\[= \frac{3}{1 - b}\]

\[\frac{b + 2}{(b - 1)^{2}} \cdot \frac{3b - 3}{b^{2} - 4} - \frac{3}{b - 2} =\]

\[= \frac{3}{1 - b}\]

\[= \frac{3}{1 - b}\]

\[\frac{3}{(b - 1)(b - 2)} - \frac{3^{\backslash b - 1}}{b - 2} = \frac{3}{1 - b}\]

\[\frac{3 - 3 \cdot (b - 1)}{(b - 1)(b - 2)} = \frac{3}{1 - b}\]

\[\frac{3 - 3b + 3}{(b - 1)(b - 2)} = \frac{3}{1 - b}\]

\[\frac{6 - 3b}{(b - 1)(b - 2)} = \frac{3}{1 - b}\]