677* Вправа 677Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 197

677. Спростіть вираз:

1) ${\displaystyle \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\frac{2\sqrt{xy}}{x-y}=}$${\displaystyle \frac{x-\sqrt{xy}+\sqrt{xy}+y+2\sqrt{xy}}{x-y}=}$${\displaystyle \frac{{(\sqrt{x}+\sqrt{y})}^2}{x-y}=}$${\displaystyle \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}}$;

${\displaystyle -\frac{\sqrt{y}+\sqrt{x}}{y-\sqrt{yx}}=}$${\displaystyle -\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}(\sqrt{y}-\sqrt{x})}=}$${\displaystyle \frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}(\sqrt{x}-\sqrt{y})}}$;

${\displaystyle \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}:\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}(\sqrt{x}-\sqrt{y})}=}$${\displaystyle \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\cdot \frac{\sqrt{y}(\sqrt{x}-\sqrt{y})}{\sqrt{x}+\sqrt{y}}=}$${\displaystyle \sqrt{y}}$;

2) ${\displaystyle \frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}}=}$${\displaystyle \frac{x+\sqrt{xy}-\sqrt{xy}+y}{x-y}=}$${\displaystyle \frac{x+y}{x-y}}$;

${\displaystyle \frac{x+y}{x-y}\cdot \frac{\sqrt{y}(x-y)}{y}=}$${\displaystyle \frac{(x+y)\sqrt{y}}{y}=}$${\displaystyle \frac{x+y}{\sqrt{y}}}$;

3) ${\displaystyle \frac{x+1+\sqrt{x^2-1}}{x+1-\sqrt{x^2-1}}=}$${\displaystyle \frac{{(x+1+\sqrt{x^2-1})}^2}{{(x+1)}^2-(x^2-1)}=}$${\displaystyle \frac{2x^2+2x+2(x+1)\sqrt{x^2-1}}{2x+2}=}$${\displaystyle \frac{2(x+1)(x+\sqrt{x^2-1})}{2(x+1)}=}$${\displaystyle x+\sqrt{x^2-1}}$;

${\displaystyle {(x+\sqrt{x^2-1}-x)}^2=}$${\displaystyle {\left(\sqrt{x^2-1}\right)}^2=}$${\displaystyle x^2-1}$;

4) ${\displaystyle \frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=}$${\displaystyle x-\sqrt{xy}+y}$;

${\displaystyle \frac{3\sqrt{xy}-3y}{x-y}=}$${\displaystyle \frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}}$;

${\displaystyle \frac{x-\sqrt{xy}+y}{x-y}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\frac{3y}{x-y}=}$${\displaystyle \frac{x-\sqrt{xy}+y+3\sqrt{y}(\sqrt{x}-\sqrt{y})+3y}{x-y}=}$${\displaystyle \frac{x+2\sqrt{xy}+y}{x-y}=}$${\displaystyle \frac{{(\sqrt{x}+\sqrt{y})}^2}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=}$${\displaystyle \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}}$.