e^2k+lambda+|2kappa+lambda-1|+e^k-lambda-1+|kappa-lambda-2|=4 шешу

$$e ^ { 2 k + \lambda } + | 2 \kappa + \lambda - 1 | + e ^ { k - \lambda - 1 } + | \kappa - \lambda - 2 | = 4$$

κ мәнін табыңыз

$$\left\{\begin{matrix}\kappa =\frac{7-e^{k-\lambda -1}-e^{\lambda +2k}}{3}\text{, }&-\frac{e^{2k+\lambda }}{3}-\frac{e^{k-\lambda -1}}{3}+\frac{\lambda }{2}+\frac{11}{6}\geq 0\text{ and }-\frac{e^{2k+\lambda }}{3}-\frac{e^{k-\lambda -1}}{3}-\lambda +\frac{1}{3}\geq 0\\\kappa =3-e^{k-\lambda -1}-e^{\lambda +2k}-2\lambda \text{, }&-e^{2k+\lambda }-e^{k-\lambda -1}-\frac{3\lambda }{2}+\frac{5}{2}\geq 0\text{ and }1-e^{k-\lambda -1}-e^{\lambda +2k}-3\lambda \leq 0\\\kappa =-2\lambda +e^{\lambda +2k}+e^{k-\lambda -1}-5\text{, }&-3\lambda +e^{\lambda +2k}+e^{k-\lambda -1}-7\geq 0\text{ and }e^{2k+\lambda }+e^{k-\lambda -1}-\frac{3\lambda }{2}-\frac{11}{2}\leq 0\\\kappa =\frac{e^{\lambda +2k}+e^{k-\lambda -1}-1}{3}\text{, }&\frac{e^{2k+\lambda }}{3}+\frac{e^{k-\lambda -1}}{3}+\frac{\lambda }{2}-\frac{5}{6}\leq 0\text{ and }\frac{e^{2k+\lambda }}{3}+\frac{e^{k-\lambda -1}}{3}-\lambda -\frac{7}{3}\leq 0\end{matrix}\right.$$