(x-y+1)(x^2+y^2-5)>0 ઉકેલો

$$( x - y + 1 ) ( x ^ { 2 } + y ^ { 2 } - 5 ) > 0$$

x માટે ઉકેલો

$$\left\{\begin{matrix}x\in \left(-\sqrt{5-y^{2}},\sqrt{5-y^{2}}\right)\text{, }&y\geq 2\text{ and }y<\sqrt{5}\\x\in \left(-\sqrt{5-y^{2}},y-1\right)\text{, }&\left(y>-1\text{ and }y<2\text{ and }|y|<\sqrt{5}\right)\text{ or }\left(y\geq 1\text{ and }y<2\right)\\x>y-1\text{, }&\left(y\geq 2\text{ and }y\leq \sqrt{5}\right)\text{ or }|y|>\sqrt{5}\\x\in \left(y-1,-\sqrt{5-y^{2}}\right)\text{, }&y\geq -\sqrt{5}\text{ and }y<-1\\x>\sqrt{5-y^{2}}\text{, }&\left(y>-1\text{ and }y<2\text{ and }|y|\leq \sqrt{5}\right)\text{ or }\left(y\geq -\sqrt{5}\text{ and }y<-1\right)\text{ or }\left(y<1\text{ and }|y|\leq \sqrt{5}\right)\text{ or }\left(y\leq 1\text{ and }y>-\sqrt{5}\text{ and }|y|\leq \sqrt{5}\right)\end{matrix}\right.$$

y માટે ઉકેલો

$$\left\{\begin{matrix}y\in \left(-\sqrt{5-x^{2}},\sqrt{5-x^{2}}\right)\text{, }&x>-\sqrt{5}\text{ and }x\leq -2\\y\in \left(x+1,\sqrt{5-x^{2}}\right)\text{, }&\left(x>-2\text{ and }x<1\text{ and }|x|<\sqrt{5}\right)\text{ or }\left(x>-2\text{ and }x\leq -1\right)\\y<x+1\text{, }&\left(x\geq -\sqrt{5}\text{ and }x\leq -2\right)\text{ or }|x|>\sqrt{5}\\y\in \left(\sqrt{5-x^{2}},x+1\right)\text{, }&x>1\text{ and }x\leq \sqrt{5}\\y<-\sqrt{5-x^{2}}\text{, }&\left(x>-2\text{ and }x<1\text{ and }|x|\leq \sqrt{5}\right)\text{ or }\left(x>1\text{ and }x\leq \sqrt{5}\right)\text{ or }\left(x>-1\text{ and }|x|\leq \sqrt{5}\right)\text{ or }\left(x\geq -1\text{ and }x<\sqrt{5}\text{ and }|x|\leq \sqrt{5}\right)\end{matrix}\right.$$