ابتدا طرفین معادله را به توان 2 می‌رسانیم: $\begin{align}  & {{(\sqrt{x+1}-\sqrt{2x-5})}^{2}}=1\Rightarrow x+1+2x-5-2\sqrt{(x+1)(2x-5)}=1 \\  & \Rightarrow {{(3x-5)}^{2}}={{\left( 2\sqrt{(x+1)(2x-5)} \right)}^{2}}\Rightarrow 9{{x}^{2}}-30x+25=8{{x}^{2}}-12x-20 \\  & \Rightarrow {{x}^{2}}-18x+45=0 \\  & \Rightarrow (x-3)(x-15)=0\Rightarrow \left\{ \begin{matrix}   x=3\Rightarrow k=3  \\   x=15  \\ \end{matrix} \right. \\  & \sqrt{x+k}=k\xrightarrow{k=3}\sqrt{x+3}=3\Rightarrow x+3=9\Rightarrow x=6 \\ \end{align}$