y=4x^4+4/5quad(y+1)^2/y=? সমাধান কৰক

$$y = \frac { 4 x ^ { 4 } + 4 } { 5 } \quad \frac { ( y + 1 ) ^ { 2 } } { y } = ?$$

x-ৰ বাবে সমাধান কৰক (জটিল সমাধান)

$$x=\left(-\frac{1}{2}+\frac{1}{2}i\right)\left(y+1\right)^{-\frac{1}{2}}\sqrt[4]{4+8y-y^{2}}$$

$$x=\left(\frac{1}{2}+\frac{1}{2}i\right)\left(y+1\right)^{-\frac{1}{2}}\sqrt[4]{4+8y-y^{2}}$$

$$x=\left(-\frac{1}{2}-\frac{1}{2}i\right)\left(y+1\right)^{-\frac{1}{2}}\sqrt[4]{4+8y-y^{2}}$$

$$x=\left(\frac{1}{2}-\frac{1}{2}i\right)\left(y+1\right)^{-\frac{1}{2}}\sqrt[4]{4+8y-y^{2}}\text{, }y\neq -1\text{ and }y\neq 0$$

y-ৰ বাবে সমাধান কৰক (জটিল সমাধান)

$$\left\{\begin{matrix}y=\frac{2\left(2x^{4}+\sqrt{5\left(x^{4}+1\right)}+2\right)}{1-4x^{4}}\text{; }y=\frac{2\left(2x^{4}-\sqrt{5\left(x^{4}+1\right)}+2\right)}{1-4x^{4}}\text{, }&x\neq \frac{\sqrt{2}}{2}\text{ and }x\neq -\frac{\sqrt{2}i}{2}\text{ and }x\neq \frac{\sqrt{2}i}{2}\text{ and }x\neq -\frac{\sqrt{2}}{2}\text{ and }x\neq \sqrt{2}\left(-\frac{1}{2}-\frac{1}{2}i\right)\text{ and }x\neq \sqrt{2}\left(\frac{1}{2}+\frac{1}{2}i\right)\text{ and }x\neq \sqrt{2}\left(-\frac{1}{2}+\frac{1}{2}i\right)\text{ and }x\neq \sqrt{2}\left(\frac{1}{2}-\frac{1}{2}i\right)\\y=-\frac{1}{2}\text{, }&x=-\frac{\sqrt{2}}{2}\text{ or }x=\frac{\sqrt{2}}{2}\text{ or }x=-\frac{\sqrt{2}i}{2}\text{ or }x=\frac{\sqrt{2}i}{2}\end{matrix}\right.$$

x-ৰ বাবে সমাধান কৰক

$$\left\{\begin{matrix}x=\frac{\sqrt{-\frac{2\sqrt{y^{2}-8y-4}}{y+1}}}{2}\text{; }x=-\frac{\sqrt{-\frac{2\sqrt{y^{2}-8y-4}}{y+1}}}{2}\text{, }&y<-1\text{ or }y=2\sqrt{5}+4\text{ or }y=4-2\sqrt{5}\\x=\frac{\sqrt{\frac{2}{y+1}}\sqrt[4]{y^{2}-8y-4}}{2}\text{; }x=-\frac{\sqrt{\frac{2}{y+1}}\sqrt[4]{y^{2}-8y-4}}{2}\text{, }&y\geq 2\sqrt{5}+4\text{ or }\left(y>-1\text{ and }y\leq 4-2\sqrt{5}\right)\end{matrix}\right.$$

y-ৰ বাবে সমাধান কৰক

$$\left\{\begin{matrix}y=\frac{2\left(2x^{4}+\sqrt{5\left(x^{4}+1\right)}+2\right)}{1-4x^{4}}\text{; }y=\frac{2\left(2x^{4}-\sqrt{5\left(x^{4}+1\right)}+2\right)}{1-4x^{4}}\text{, }&|x|\neq \frac{\sqrt{2}}{2}\\y=-\frac{1}{2}\text{, }&|x|=\frac{\sqrt{2}}{2}\end{matrix}\right.$$