The solution of $x-1=\left(x-\left[x\right]\right)\left(x-\left\{x\right\}\right)$  (where $\left[x\right]$  and $\left\{x\right\}$  are the integral and fractional part of $x$ ) is :

$\because$            $\left(x-1\right)=\left(x-\left[x\right]\right)\left(x-\left\{x\right\}\right)$

$\Rightarrow$            $x=1+\left\{x\right\}\left[x\right]$

$\Rightarrow$          $\left[x\right]+\left\{x\right\}=1+\left\{x\right\}\left[x\right]$

$\Rightarrow$          $\left(\left\{x\right\}-1\right)\left(\left[x\right]-1\right)=0$

         $\left\{x\right\}-1\ne 0,\therefore \left[x\right]-1=0$

               $\left[x\right]=1\Rightarrow x\in [1,2)$