Задача 79938 Номер 15 и 16 решить ...

6654737b3528112db82d8cf9

6/2/2025 5:04:58 PM

Номер 15 и 16 решить

626fab9a354ab65fb2f43abb

6/2/2025 9:13:19 PM

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15) [m]\large (\frac{ax^3 + bx^2 + c}{(a+b)x})' = \frac{(3ax^2+2bx)(a+b)x - (ax^3 + bx^2 + c)(a+b)}{(a+b)^2x^2} = \frac{(3ax^2+2bx)x - (ax^3 + bx^2 + c)}{(a+b)x^2} =[/m]
[m]\large = \frac{3ax^3+2bx^2 - ax^3 - bx^2 - c}{(a+b)x^2} = \frac{2ax^3+bx^2 - c}{(a+b)x^2}[/m]

16) [m]\large \bigg ( \left ( \frac{mu + n}{p} \right )^3 \bigg )' = \bigg ( \frac{(mu + n)^3}{p^3} \bigg )' = \frac{3(mu + n)^2 \cdot (mu + n)'}{p^3} = \frac{3m(mu + n)^2}{p^3}[/m]