Тип 8 № 311758 Найдите значение выражения \(\frac{4 + \frac{b}{4}}{\frac{b}{64b^2 + 128b + 64}\) при b = \(\frac{15}{16}\)

Сначала упростим выражение: \(\frac{4 + \frac{b}{4}}{\frac{b}{64b^2 + 128b + 64}} = \frac{4 + \frac{b}{4}}{ \frac{b}{64(b^2 + 2b + 1)}} = \frac{4 + \frac{b}{4}}{\frac{b}{64(b+1)^2}} = (4 + \frac{b}{4}) * \frac{64(b+1)^2}{b}\) Теперь подставим значение b = \(\frac{15}{16}\): \((4 + \frac{\frac{15}{16}}{4}) * \frac{64(\frac{15}{16}+1)^2}{\frac{15}{16}} = (4 + \frac{15}{64}) * \frac{64(\frac{31}{16})^2}{\frac{15}{16}} = (\frac{256+15}{64}) * \frac{64 * \frac{961}{256}}{\frac{15}{16}} = \frac{271}{64} * \frac{\frac{961}{4}}{\frac{15}{16}} = \frac{271}{64} * \frac{961}{4} * \frac{16}{15} = \frac{271}{4} * \frac{961}{15} = \frac{260431}{60}\) ≈ 4340.51 Ответ: \(\frac{260431}{60}\)