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:

\[1)\ \frac{b^{3}}{b^{2} - 8b + 16} - \frac{{b^{2}}^{\backslash b - 4}}{b - 4} =\]

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

\[2)\ \frac{b^{2}}{b^{2} - 16} - \frac{b^{\backslash b + 4}}{b - 4} =\]

\[= \frac{b^{2} - b^{2} - 4b}{(b - 4)(b + 4)} = - \frac{4b}{b^{2} - 16};\]

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

\[= - \frac{4b^{2}}{(b - 4)^{2}} \cdot \frac{b^{2} - 16}{4b} =\]

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

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

\[Что\ и\ требовалось\ доказать.\]