упростите с решением , пожалуйста [tex] \sqrt{x-2 \sqrt{x-1}} - \sqrt{x+2 \sqrt{x-1} } [/tex]. Created by Pomochnik1vsegda. algebra-ru

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с решением , пожалуйста [tex] \sqrt{x-2 \sqrt{x-1}} - \sqrt{x+2 \sqrt{x-1} } [/tex]

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SYSTEMCORE

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$\displaystyle \sqrt{x-2\sqrt{x-1}}-\sqrt{x+2\sqrt{x-1}}=\sqrt{(x-1)-2\sqrt{x-1}+1}-\\\\\\-\sqrt{(x-1)+2\sqrt{x-1}+1}=\sqrt{(\sqrt{x-1}-1)^2}-\sqrt{(\sqrt{x-1}+1)^2}=\\\\\\=|\sqrt{x-1}-1|-|\sqrt{x-1}+1|\\\\\\1)\,\,\,\sqrt{x-1}-1\ \textgreater \ 0\\\\\sqrt{x-1}\ \textgreater \ 1\\\\x-1\ \textgreater \ 1\\\\x\ \textgreater \ 2\\\\\sqrt{x-1}-1-(\sqrt{x-1}-1)=-1-1=-2\\\\\\2)\sqrt{x-1}-1 \leq 0\\\\\sqrt{x-1} \leq 1\\\\x-1 \leq 1\\\\x \leq 2\\\\\text{ODZ}:x \geq 1\quad \rightarrow \quad x\in[1;2]\\\\-\sqrt{x-1}+1-(\sqrt{x-1}-1)=-2\sqrt{x-1}$

$|\sqrt{x-1}-1|-|\sqrt{x-1}+1|=\boxed{ \left \{ {{-2\sqrt{x-1},\,\,\,x\in[1;2]} \atop {-2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\ \textgreater \ 2}} \right. }$

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