What is the value of x in the equation 1 + xx - x1 + x = 1 6 ?

Answer: Option C

$$\begin{array}{rl}& \sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}=\frac{1}{\sqrt{6}}\\ & \text{On squaring both sides,}\\ & \Rightarrow {\left(\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}\right)}^2=\frac{1}{6}\\ & \Rightarrow \frac{1+x}{x}+\frac{x}{1+x}-2=\frac{1}{6}\\ & \Rightarrow \frac{1+x}{x}+\frac{x}{1+x}=2+\frac{1}{6}\\ & \Rightarrow \frac{1+x}{x}+\frac{x}{1+x}=\frac{13}{6}\\ & \Rightarrow \frac{{\left(1+x\right)}^2+x^2}{x\left(1+x\right)}=\frac{13}{6}\\ & \Rightarrow \frac{1+x^2+2x+x^2}{x+x^2}=\frac{13}{6}\\ & \Rightarrow \left(2x^2+2x+1\right)6=13\left(x+x^2\right)\\ & \Rightarrow 12x^2+12x+6=13x+13x^2\\ & \Rightarrow x^2+x-6=0\\ & \Rightarrow x^2+3x-2x-6=0\\ & \Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\\ & \Rightarrow \left(x+3\right)\left(x-2\right)=0\\ & \Rightarrow x=2\text{ as }x\ne -3\\ & \text{Or, of the given options, when }x=2\\ & \text{L}\text{.H}\text{.S}\text{.}=\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}\\ & =\sqrt{\frac{1+2}{2}}-\sqrt{\frac{2}{1+2}}\\ & =\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}}\\ & =\frac{3-2}{\sqrt{6}}\\ & =\frac{1}{\sqrt{6}}\end{array}$$