Question

Помогите срочно . Даю 20 балов за решение этой задачи!!!

Answer 1

$\boxed {(\sqrt{x})^2=\sqrt{x}\cdot \sqrt{x}=x}\\\\\\\Big ( \frac{\sqrt{x}-1}{\sqrt{x}+1} +\frac{2\sqrt{x}}{x-1} \Big ): \frac{\sqrt{x}+1}{x-\sqrt{x}} =\\\\=\Big ( \frac{\sqrt{x}-1}{\sqrt{x}+1} + \frac{2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)} \Big ): \frac{\sqrt{x}+1}{\sqrt{x}\cdot (\sqrt{x}-1)} =$

$=\frac{(\sqrt{x}-1)^2+2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\cdot \frac{\sqrt{x}\cdot (\sqrt{x}-1))}{\sqrt{x}+1} = \frac{(x-2\sqrt{x}+1+2\sqrt{x})\cdot \sqrt{x}}{(\sqrt{x}+1)^2} = \frac{\sqrt{x}\cdot (x+1)}{(\sqrt{x}+1)^2}$