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Решите задачу:

$\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}} =\\\\=\sqrt{(1+\sqrt{x-1})^2}+\sqrt{(1-\sqrt{x-1})^2} =\\\\=|1+\sqrt{x-1}|+|1-\sqrt{x-1}|=\\\\=1+\sqrt{x-1}+|1-\sqrt{x-1}|=A\\\\a)\; \; \; esli\; \; (1-\sqrt{x-1})\geq 0\; ,\; to\; \; |1-\sqrt{x-1}|=1-\sqrt{x-1},\; togda\\\\A=1+\sqrt{x-1}+1-\sqrt{x-1}=2\; ;\\\\\star \; 1-\sqrt{x-1}\geq 0\; ,\; \; \sqrt{x-1}\leq 1\; ,\; \; x-1\leq 1\; ,\; x\leq 2\\\\\star \; \; x-1\geq 0\; ,\; \; x\geq 1\\\\\underline {1\leq x\leq 2}\\\\b)\; \; esli\; \; (1-\sqrt{x-1})\ \textless \ 0\; ,\; to\; \; |1-\sqrt{x-1}|=\sqrt{x-1}-1.\; togda\\\\A=1+\sqrt{x-1}+\sqrt{x-1}-1=2\sqrt{x-1}\; .$

$\star \; \; 1-\sqrt{x-1}\ \textless \ 0\; ,\; \; \sqrt{x-1}\ \textgreater \ 1\; ,\; \; x-1\ \textgreater \ 1\; ,\; \; \underline {x\ \textgreater \ 2}\\\\c)\; \; A= \left \{ {{2\; ,\; esli\; \; 1 \leq x \leq 2} \atop {2\sqrt{x-1}\; ,\; esli\; \; x\ \textgreater \ 2}} \right.$