Question

Решите пожалуйста 1 пример!! Даю 20 баллов!
$\sqrt{x+2\sqrt{x-1}} - \sqrt{x-2\sqrt{x-1}};$

Answer 1

$\sqrt{x+2\sqrt{x-1}} -\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1+2\sqrt{x-1}+1} -\\\\-\sqrt{x-1-2\sqrt{x-1}+1}= \sqrt{(\sqrt{x-1})^2+2\cdot\sqrt{x-1}\cdot 1+1^2} -\\ \\ - \sqrt{(\sqrt{x-1})^2-2\cdot\sqrt{x-1}\cdot 1+1^2}=\sqrt{(\sqrt{x-1}+1)^2} -\\\\ -\sqrt{(\sqrt{x-1}-1)^2} =|{\sqrt{x-1}+1}|-|{\sqrt{x-1}-1}|=\\\\=\sqrt{x-1}+1-|{\sqrt{x-1}-1}|=\left \{ {{\sqrt{x-1}+1-(\sqrt{x-1}-1), \sqrt{x-1}-1\geq0} \atop {{\sqrt{x-1}+1-(1-\sqrt{x-1}), \sqrt{x-1}-1<0}} \right. =$

$=\left \{ {{\sqrt{x-1}+1-\sqrt{x-1}+1, \sqrt{x-1}\geq1} \atop {{\sqrt{x-1}+1-1+\sqrt{x-1}, \sqrt{x-1}<1}} \right.=\left \{ {{2, x-1\geq1} \atop {{2\sqrt{x-1}, 0\leq x-1<1}} \right. =\left \{ {{2, x\geq2} \atop {{2\sqrt{x-1}, 1\leq x<2}} \right.$