I need answer

$$\begin{gathered} Dear Student \\ i) \\ \frac{1}{x}-\frac{1}{x+1} =\frac{1}{6} \\ 6\left((x+1) -x\right) =(x+1\left)x \\ 6\right(x+1-x) =x^2+x \\ 6=x^2+x \\ {x}^2+x-6 =0 \\ {x}^2+3x-2x-6 = 0 \\ x(x+3) -2(x+3) =0 \\ (x+3\left)\right(x-2) =0 \\ x =-3 \\ or \\ x =2 \end{gathered}$$

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