Решить 1/r+nur=v^2quadr

$$\frac { 1 } { r + \nu r } = v ^ { 2 } \quad r$$

Найдите r (комплексное решение)

$$r=-\frac{\left(\nu +1\right)^{-\frac{1}{2}}}{v}$$

$$r=\frac{\left(\nu +1\right)^{-\frac{1}{2}}}{v}\text{, }v\neq 0\text{ and }\nu \neq -1$$

Найдите v (комплексное решение)

$$v=-\frac{\left(\nu +1\right)^{-\frac{1}{2}}}{r}$$

$$v=\frac{\left(\nu +1\right)^{-\frac{1}{2}}}{r}\text{, }r\neq 0\text{ and }\nu \neq -1$$

Найдите r

$$r=\frac{1}{\sqrt{\nu +1}v}$$

$$r=-\frac{1}{\sqrt{\nu +1}v}\text{, }v\neq 0\text{ and }\nu >-1$$

Найдите v

$$v=\frac{1}{\sqrt{\nu +1}r}$$

$$v=-\frac{1}{\sqrt{\nu +1}r}\text{, }r\neq 0\text{ and }\nu >-1$$