פתור את 32^y/x^2+16/x=5y/x+5

$$\frac { 32 ^ { y } } { x ^ { 2 } } + \frac { 16 } { x } = \frac { 5 y } { x } + 5$$

פתור עבור x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{25y^{2}-160y+20\times 32^{y}+256}}{10}-\frac{y}{2}+\frac{8}{5}\text{, }&\frac{\sqrt{20\times 32^{y}+\left(16-5y\right)^{2}}}{10}-\frac{y}{2}+\frac{8}{5}\neq 0\\x=-\frac{\sqrt{25y^{2}-160y+20\times 32^{y}+256}}{10}-\frac{y}{2}+\frac{8}{5}\text{, }&-\frac{\sqrt{20\times 32^{y}+\left(16-5y\right)^{2}}}{10}-\frac{y}{2}+\frac{8}{5}\neq 0\end{matrix}\right.$$

פתור עבור x

$$\left\{\begin{matrix}x=\frac{\sqrt{25y^{2}-160y+20\times 32^{y}+256}}{10}-\frac{y}{2}+\frac{8}{5}\text{, }&\frac{\sqrt{25y^{2}-160y+20\times 32^{y}+256}}{10}-\frac{y}{2}+\frac{8}{5}\neq 0\\x=-\frac{\sqrt{25y^{2}-160y+20\times 32^{y}+256}}{10}-\frac{y}{2}+\frac{8}{5}\text{, }&-\frac{\sqrt{25y^{2}-160y+20\times 32^{y}+256}}{10}-\frac{y}{2}+\frac{8}{5}\neq 0\end{matrix}\right.$$