Ответы
Ответ дал:
1
Применены : свойства логарифмов, формула суммы кубов
Приложения:
![](https://st.uroker.com/files/e0d/e0dee82b891bf9f6e2ae9f31cc5024ed.jpg)
Ответ дал:
1
разложение многочлена по формуле кубов суммы и далее свойства логарифмов
![\frac{log _{4}(b^2-3b+9)-(log _{4}(b+3) +log _{4}(b^2-3b+9) }{64log _{4}(b+3)} \frac{log _{4}(b^2-3b+9)-(log _{4}(b+3) +log _{4}(b^2-3b+9) }{64log _{4}(b+3)}](https://tex.z-dn.net/?f=+%5Cfrac%7Blog+_%7B4%7D%28b%5E2-3b%2B9%29-%28log+_%7B4%7D%28b%2B3%29+%2Blog+_%7B4%7D%28b%5E2-3b%2B9%29+%7D%7B64log+_%7B4%7D%28b%2B3%29%7D+)
![\frac{log _{4} (b^2-6b+3b+9)-log _{4} (b+3)(b^2-3b+9)}{64log _{4} (b+3)} \frac{log _{4} (b^2-6b+3b+9)-log _{4} (b+3)(b^2-3b+9)}{64log _{4} (b+3)}](https://tex.z-dn.net/?f=+%5Cfrac%7Blog++_%7B4%7D++%28b%5E2-6b%2B3b%2B9%29-log+_%7B4%7D+%28b%2B3%29%28b%5E2-3b%2B9%29%7D%7B64log+_%7B4%7D+%28b%2B3%29%7D+)
![\frac{log _{4} (b^2-3b+9) - (log _{4}(b+3)+log _{4}(b^2-3b+9))}{64log _{4}(b+3)} \frac{log _{4} (b^2-3b+9) - (log _{4}(b+3)+log _{4}(b^2-3b+9))}{64log _{4}(b+3)}](https://tex.z-dn.net/?f=+%5Cfrac%7Blog+_%7B4%7D+%28b%5E2-3b%2B9%29+-+%28log+_%7B4%7D%28b%2B3%29%2Blog+_%7B4%7D%28b%5E2-3b%2B9%29%29%7D%7B64log+_%7B4%7D%28b%2B3%29%7D+)
![-\frac{log _{4} (b+3) }{64log _{4}(b+3)}* 5 ^{2log _{5}8} = \frac{-1}{64} *5 ^{log _{5}8^2} -\frac{log _{4} (b+3) }{64log _{4}(b+3)}* 5 ^{2log _{5}8} = \frac{-1}{64} *5 ^{log _{5}8^2}](https://tex.z-dn.net/?f=+-%5Cfrac%7Blog+_%7B4%7D+%28b%2B3%29+%7D%7B64log+_%7B4%7D%28b%2B3%29%7D%2A+5+%5E%7B2log+_%7B5%7D8%7D+%3D++%5Cfrac%7B-1%7D%7B64%7D+%2A5+%5E%7Blog+_%7B5%7D8%5E2%7D+)
![\frac{-1}{64} * 8^2 = - \frac{1}{64} *64 = -1 \frac{-1}{64} * 8^2 = - \frac{1}{64} *64 = -1](https://tex.z-dn.net/?f=+%5Cfrac%7B-1%7D%7B64%7D+%2A+8%5E2+%3D+-+%5Cfrac%7B1%7D%7B64%7D+%2A64+%3D+-1)
Вас заинтересует
4 месяца назад
4 месяца назад
1 год назад
1 год назад
2 года назад