n(x+y)+m(x-y)=m^2-mn+n^2 шешу

$$n ( x + y ) + m ( x - y ) = m ^ { 2 } - m n + n ^ { 2 }$$

m мәнін табыңыз (complex solution)

$$m=\frac{-\sqrt{x^{2}+6nx-2xy+y^{2}+2ny-3n^{2}}+n+x-y}{2}$$

$$m=\frac{\sqrt{x^{2}+6nx-2xy+y^{2}+2ny-3n^{2}}+n+x-y}{2}$$

n мәнін табыңыз (complex solution)

$$n=\frac{-\sqrt{x^{2}+2xy+6mx+y^{2}-2my-3m^{2}}+m+x+y}{2}$$

$$n=\frac{\sqrt{x^{2}+2xy+6mx+y^{2}-2my-3m^{2}}+m+x+y}{2}$$

m мәнін табыңыз

$$m=\frac{-\sqrt{x^{2}+6nx-2xy+y^{2}+2ny-3n^{2}}+n+x-y}{2}$$

$$m=\frac{\sqrt{x^{2}+6nx-2xy+y^{2}+2ny-3n^{2}}+n+x-y}{2}\text{, }\left(n=0\text{ or }x\neq 0\text{ or }y\neq 0\right)\text{ and }\left(x\neq 0\text{ or }n\geq \frac{y-2|y|}{3}\right)\text{ and }\left(x\neq 0\text{ or }n\leq \frac{2|y|+y}{3}\right)\text{ and }\left(n=0\text{ or }n\geq -\frac{2\sqrt{3x^{2}+y^{2}}}{3}+\frac{y}{3}+x\right)\text{ and }\left(n=0\text{ or }n\leq \frac{2\sqrt{3x^{2}+y^{2}}}{3}+\frac{y}{3}+x\right)$$

n мәнін табыңыз

$$n=\frac{-\sqrt{x^{2}+2xy+6mx+y^{2}-2my-3m^{2}}+m+x+y}{2}$$

$$n=\frac{\sqrt{x^{2}+2xy+6mx+y^{2}-2my-3m^{2}}+m+x+y}{2}\text{, }y=-\frac{3m}{2}\text{ or }\left(m>0\text{ and }y\leq -\frac{3m}{2}\right)\text{ or }\left(y\geq -\frac{3m}{2}\text{ and }m<0\right)\text{ or }x\geq 2\sqrt{2my+3m^{2}}-y-3m\text{ or }x\leq -2\sqrt{2my+3m^{2}}-y-3m$$