सोडोवचें y=frac{sqrt{y}+1}{sqrt{x}}

$$y = \frac { \sqrt { y } + 1 } { \sqrt { x } }$$

x खातीर सोडोवचें (जटील सोल्यूशन)

$$x=\left(y^{-\frac{1}{2}}+\frac{1}{y}\right)^{2}$$

$$arg(y^{-\frac{1}{2}}+\frac{1}{y})<\pi \text{ and }y\neq 0$$

y खातीर सोडोवचें (जटील सोल्यूशन)

$$\left\{\begin{matrix}y=\frac{2\sqrt{x}+\sqrt{4\sqrt{x}+1}+1}{2x}\text{, }&x\neq 0\text{ and }arg(\frac{x^{-\frac{1}{2}}+x^{-\frac{1}{2}}\sqrt{4\sqrt{x}+1}}{2})<\pi \\y=\frac{2\sqrt{x}-\sqrt{4\sqrt{x}+1}+1}{2x}\text{, }&x\neq 0\text{ and }arg(\frac{x^{-\frac{1}{2}}-x^{-\frac{1}{2}}\sqrt{4\sqrt{x}+1}}{2})<\pi \end{matrix}\right.$$

x खातीर सोडोवचें

$$x=\left(\frac{1}{\sqrt{y}}+\frac{1}{y}\right)^{2}$$

$$y>0$$

y खातीर सोडोवचें

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

$$x\geq 4$$