Ответ:
![\Large \displaystyle y = \ln(1 + x) - 1 + \frac{\ln(1 + x)}{x} + \frac{C_3}{x}](https://tex.z-dn.net/?f=%5CLarge+%5Cdisplaystyle+y+%3D+%5Cln%281+%2B+x%29+-+1+%2B+%5Cfrac%7B%5Cln%281+%2B+x%29%7D%7Bx%7D+%2B+%5Cfrac%7BC_3%7D%7Bx%7D)
Пошаговое объяснение:
![y' + \frac{y}{x} = \frac{\ln(1 + x)}{x}\\xy' + y = \ln(1 + x)\\x\frac{dy}{dx} + y = \ln(1 + x)](https://tex.z-dn.net/?f=y%27+%2B+%5Cfrac%7By%7D%7Bx%7D+%3D+%5Cfrac%7B%5Cln%281+%2B+x%29%7D%7Bx%7D%5C%5Cxy%27+%2B+y+%3D+%5Cln%281+%2B+x%29%5C%5Cx%5Cfrac%7Bdy%7D%7Bdx%7D+%2B+y+%3D+%5Cln%281+%2B+x%29)
По формуле производной произведения:
, так как
.
![\frac{d(xy)}{dx} = \ln(1 + x)\\\int {d(xy)} = \int {\ln(1 + x)} \, dx\\xy + C_1= \int {\ln(1 + x)} \, dx](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28xy%29%7D%7Bdx%7D+%3D+%5Cln%281+%2B+x%29%5C%5C%5Cint+%7Bd%28xy%29%7D+%3D+%5Cint+%7B%5Cln%281+%2B+x%29%7D+%5C%2C+dx%5C%5Cxy+%2B+C_1%3D+%5Cint+%7B%5Cln%281+%2B+x%29%7D+%5C%2C+dx)
Интегрируем по частям:
![\int {\ln(1 + x)} \, dx = x\ln(1 + x) - \int {x} \, d(\ln(1 + x)) = x\ln(1 + x) - \int {(1 - \frac{1}{x + 1})} \, dx =\\= x\ln(1 + x) - x + \ln(1 + x) + C_2](https://tex.z-dn.net/?f=%5Cint+%7B%5Cln%281+%2B+x%29%7D+%5C%2C+dx+%3D+x%5Cln%281+%2B+x%29+-+%5Cint+%7Bx%7D+%5C%2C+d%28%5Cln%281+%2B+x%29%29+%3D+x%5Cln%281+%2B+x%29+-+%5Cint+%7B%281+-+%5Cfrac%7B1%7D%7Bx+%2B+1%7D%29%7D+%5C%2C+dx+%3D%5C%5C%3D+x%5Cln%281+%2B+x%29+-+x+%2B+%5Cln%281+%2B+x%29+%2B+C_2)
![xy = x\ln(1 + x) - x + \ln(1 + x) + C_3\\y = \ln(1 + x) - 1 + \frac{\ln(1 + x)}{x} + \frac{C_3}{x}](https://tex.z-dn.net/?f=xy+%3D+x%5Cln%281+%2B+x%29+-+x+%2B+%5Cln%281+%2B+x%29+%2B+C_3%5C%5Cy+%3D+%5Cln%281+%2B+x%29+-+1+%2B+%5Cfrac%7B%5Cln%281+%2B+x%29%7D%7Bx%7D+%2B+%5Cfrac%7BC_3%7D%7Bx%7D)
25:100*60=15 мальчики
25-15=10 девочки
откуда мы знаем что решать?
Пусть х - задуманное число, тогда составим и решим уравнение
(х+9,2)×11=110
решение:
х+9,2=110÷11
х+9,2=10
х=10-9,2
х=0.8
ответ: 0.8