{l}{x^2+y^2=+z^2z^2}{2lambdax+4y=0}{2dy+4x+z=0} حل کریں

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = + z ^ { 2 } z ^ { 2 } } \\ { 2 \lambda x + 4 y = 0 } \\ { 2 d y + 4 x + z = 0 } \end{array} \right.$$

x، y، z کے لئے حل کریں

$$\left\{\begin{matrix}\\x=0\text{, }y=0\text{, }z=0\text{, }&\text{unconditionally}\\x=-\frac{\sqrt{\lambda ^{2}+4}}{2\left(d\lambda -4\right)^{2}}\text{, }y=\frac{\lambda \sqrt{\lambda ^{2}+4}}{4\left(d\lambda -4\right)^{2}}\text{, }z=-\frac{\sqrt{\lambda ^{2}+4}}{2\left(d\lambda -4\right)}\text{; }x=\frac{\sqrt{\lambda ^{2}+4}}{2\left(d\lambda -4\right)^{2}}\text{, }y=-\frac{\lambda \sqrt{\lambda ^{2}+4}}{4\left(d\lambda -4\right)^{2}}\text{, }z=\frac{\sqrt{\lambda ^{2}+4}}{2\left(d\lambda -4\right)}\text{, }&d=0\text{ or }\lambda \neq \frac{4}{d}\end{matrix}\right.$$