Докажите тождество (b^3/(b^2-8b+16)-b^2/(b-4)) ∶(b^2/(b^2-16)-b/(b-4))=(b^2+4b)/(4-b).

Ответ:

\[\left( \frac{b^{3}}{b^{2} - 8b + 16} - \frac{b^{2}}{b - 4} \right)\ :\left( \frac{b^{2}}{b^{2} - 16} - \frac{b}{b - 4} \right) =\]

\[= \frac{b² + 4b}{4 - b}\]

\[\left( \frac{b^{3}}{(b - 4)^{2}} - \frac{{b^{2}}^{\backslash b - 4}}{b - 4} \right)\ :\left( \frac{b^{2}}{(b - 4)(b + 4)} - \frac{b^{\backslash b + 4}}{b - 4} \right) =\]

\[= \frac{b² + 4b}{4 - b}\]

\[\frac{b³ - b²(b - 4)}{(b - 4)²}\ :\frac{b² - b(b + 4)}{(b - 4)(b + 4)} = \frac{b² + 4b}{4 - b}\]

\[\frac{(b^{3} - b^{3} + 4b^{2})(b - 4)(b + 4)}{(b - 4)²(b^{2} - b^{2} - 4b)} = \frac{b² + 4b}{4 - b}\]

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

\[\frac{b² + 4b}{4 - b}\mathbf{=}\frac{b² + 4b}{4 - b}.\]