Упростите с решением, пожалуйста \sqrt{x-2 \sqrt{x-1}} - \sqrt{x+2 \sqrt{x-1}}

Решите задачу:

$\displaystyle \sqrt{x-2\sqrt{x-1}}-\sqrt{x+2\sqrt{x-1}}=\sqrt{(x-1)-2\sqrt{x-1}+1}-\\\\\\-\sqrt{(x-1)+2\sqrt{x-1}+1}=\sqrt{(\sqrt{x-1}-1)^2}-\sqrt{(\sqrt{x-1}+1)^2}\\\\\\=|\sqrt{x-1}-1|-|\sqrt{x-1}+1|\\\\\\1)\,\,\,\sqrt{x-1}-1\ \textgreater \ 0\\\\\sqrt{x-1}\ \textgreater \ 1\\\\x-1\ \textgreater \ 1\\\\x\ \textgreater \ 2\\\\\sqrt{x-1}-1-(\sqrt{x-1}-1)=-1-1=-2\\\\\\2)\sqrt{x-1}-1 \leq 0\\\\\sqrt{x-1} \leq 1\\\\x-1 \leq 1\\\\x \leq 2$

$\text{ODZ}:x \geq 1\quad \rightarrow \quad x\in[1;2]\\\\-\sqrt{x-1}+1-(\sqrt{x-1}-1)=-2\sqrt{x-1}\\\\\\|\sqrt{x-1}-1|-|\sqrt{x-1}+1|=\boxed{ \left \{ {{-2\sqrt{x-1},\,\,\,x\in[1;2]} \atop {-2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\ \textgreater \ 2}} \right. }$