حل sqrt{x+2}+sqrt{y-1}=x+y/4

$$\sqrt { x + 2 } + \sqrt { y - 1 } = \frac { x + y } { 4 }$$

برای x حل کنید (complex solution)

$$\left\{\begin{matrix}x=-y+4\sqrt{-y+4\sqrt{y-1}+6}+4\sqrt{y-1}+8\text{, }&arg(\sqrt{-y+4\sqrt{y-1}+6}+2)<\pi \\x=-y-4\sqrt{-y+4\sqrt{y-1}+6}+4\sqrt{y-1}+8\text{, }&y=4\sqrt{5}+10\text{ or }arg(-\sqrt{-y+4\sqrt{y-1}+6}+2)<\pi \end{matrix}\right.$$

برای y حل کنید (complex solution)

$$\left\{\begin{matrix}y=-x+4\sqrt{-x+4\sqrt{x+2}+3}+4\sqrt{x+2}+8\text{, }&arg(\frac{x}{4}+\frac{y}{4}-\sqrt{x+2})<\pi \\y=-x-4\sqrt{-x+4\sqrt{x+2}+3}+4\sqrt{x+2}+8\text{, }&x=4\sqrt{5}+7\text{ or }arg(\frac{x}{4}+\frac{y}{4}-\sqrt{x+2})<\pi \end{matrix}\right.$$

برای x حل کنید

$$\left\{\begin{matrix}x=-y+4\sqrt{-y+4\sqrt{y-1}+6}+4\sqrt{y-1}+8\text{, }&y\geq 1\text{ and }y\leq 26\\x=-y-4\sqrt{-y+4\sqrt{y-1}+6}+4\sqrt{y-1}+8\text{, }&y\geq 4\sqrt{5}+10\text{ and }y\leq 26\end{matrix}\right.$$

برای y حل کنید

$$\left\{\begin{matrix}y=-x+4\sqrt{-x+4\sqrt{x+2}+3}+4\sqrt{x+2}+8\text{, }&x\geq -2\text{ and }x\leq 23\\y=-x-4\sqrt{-x+4\sqrt{x+2}+3}+4\sqrt{x+2}+8\text{, }&x\geq 4\sqrt{5}+7\text{ and }x\leq 23\end{matrix}\right.$$