Question

Срочно! Помогите!Даю 25 баллов​

Answer 1

$\; \; y>0\\\\\Big (\sqrt{\frac{1}{4y^2}-1}-\frac{1}{2y}\Big )\cdot \Big (\frac{\sqrt{1+2y}}{\sqrt{1+2y}-\sqrt{1-2y}}+\frac{1-2y}{\sqrt{1-4y^2}+2y-1}\Big )=\\\\=\Big (\sqrt{\frac{1-4y^2}{4y^2}}-\frac{1}{2y}\Big )\cdot \Big (\frac{\sqrt{1+2y}}{\sqrt{1+2y}-\sqrt{1-2y}}}+\frac{1-2y}{\sqrt{(1-2y)(1+2y)}-(\sqrt{1-2y})^2}\Big )=$

$\frac{\sqrt{1-4y^2}\, -1}{2y}\cdot \Big (\frac{\sqrt{1+2y}}{\sqrt{1+2y}-\sqrt{1-2y}}+\frac{(\sqrt{1-2y})^2}{\sqrt{1-2y}\cdot (\sqrt{1+2y}-\sqrt{1-2y})}}\Big )=\\\\=\frac{\sqrt{1-4y^2}\, -1}{2y}\cdot \frac{\sqrt{1+2y}+\sqrt{1-2y}}{\sqrt{1-2y}-\sqrt{1+2y}}=\frac{(\sqrt{1-4y^2}\, -1)\cdot (\sqrt{1-2y}+\sqrt{1+2y})^2}{2y\cdot ((1-2y)-(1+2y))}=\\\\=\frac{(\sqrt{1-4y^2}-1)\cdot (1-2y+1+2y+2\sqrt{(1-2y)(1+2y)})}{2y\cdot (-4y)}=\\\\=\frac{(\sqrt{1-4y^2}-1)(2+2\sqrt{1-4y^2})}{-8y^2}=\frac{2\cdot (\sqrt{1-4y^2}-1)(1+\sqrt{1-4y^2})}{-8y^2}=\\\\=\frac{2\cdot (1-4y^2-1)}{-8y^2}=\frac{-8y^2}{-8y^2}=1$