{l}{n_1^+n2=54}{n_{1}n_{2}=38} шешу

$$\left. \begin{array} { l } { n _ { 1 } ^ { + n } 2 = 54 } \\ { n _ { 1 } n _ { 2 } = 38 } \end{array} \right.$$

n_1, n, n_2 мәнін табыңыз (complex solution)

$$n_{1}=\frac{38}{n_{2}}$$

$$n\in \mathrm{C}$$

$$n_{2}=e^{-\frac{2\pi n_{1}\left(Im(n)+iRe(n)\right)}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}}\times 27^{\frac{Re(n)-iIm(n)}{\left(Re(n)\right)^{2}+\left(Im(n)\right)^{2}}}$$

$$n_{1}\in \mathrm{Z}$$

n_1, n, n_2 мәнін табыңыз

$$\left\{\begin{matrix}\\n_{1}=\frac{38}{n_{2}}\text{, }n=\frac{3\ln(3)}{-\ln(n_{2})+\ln(38)}\text{, }n_{2}\in \left(0,38\right)\text{; }n_{1}=\frac{38}{n_{2}}\text{, }n=\frac{3\ln(3)}{-\ln(n_{2})+\ln(38)}\text{, }n_{2}>38\text{, }&\text{unconditionally}\\n_{1}=\frac{38}{n_{2}}\text{, }n=\frac{3\ln(3)}{3\ln(-\frac{1}{\sqrt[3]{n_{2}}})+\ln(38)}\text{, }n_{2}<-38\text{; }n_{1}=\frac{38}{n_{2}}\text{, }n=\frac{3\ln(3)}{3\ln(-\frac{1}{\sqrt[3]{n_{2}}})+\ln(38)}\text{, }n_{2}\in \left(-38,0\right)\text{, }&n_{2}\neq -38\text{ and }n_{2}<0\text{ and }Numerator(\frac{\ln(3)}{\ln(-\frac{1}{\sqrt[3]{n_{2}}})+\frac{\ln(38)}{3}})\text{bmod}2=0\text{ and }Denominator(\frac{\ln(3)}{\ln(-\frac{1}{\sqrt[3]{n_{2}}})+\frac{\ln(38)}{3}})\text{bmod}2=1\end{matrix}\right.$$