Докажите тождество (b/(b^2-8b+16)-(b+6)/(b^2-16))∶(b+12)/(b^2-16)=2/(b-4).

Ответ:

\[\left( \frac{b}{b^{2} - 8b + 16} - \frac{b + 6}{b^{2} - 16} \right)\ :\frac{b + 12}{b^{2} - 16} = \frac{2}{b - 4}\]

\[\left( \frac{b}{(b - 4)^{2}} - \frac{b + 6}{(b - 4)(b + 4)} \right)\ :\frac{b + 12}{(b - 4)(b + 4)} = \frac{2}{b - 4}\]

\[\frac{\left( b(b + 4) - (b + 6)(b - 4) \right)(b - 4)(b + 4)}{(b - 4)²(b + 4)(b + 12)} = \frac{2}{b - 4}\]

\[\frac{b² + 4b - b^{2} + 4b - 6b + 24}{(b - 4)(b + 12)} = \frac{2}{b - 4}\]

\[\frac{2b + 24}{(b - 4)(b + 12)} = \frac{2}{b - 4}\]

\[\frac{2 \cdot (b + 12)}{(b - 4)(b + 12)} = \frac{2}{b - 4}\]

\[\frac{2}{b - 4} = \frac{2}{b - 4}\]

\[Что\ и\ требовалось\ доказать.\]