Question

Я ВАС ОЧЕНЬ ПРОШУ, ПОМОГИТЕ! ДАМ 40 БАЛЛОВ​

Answer 1

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

$P.S.\; \; \; \sqrt{x}+y\sqrt{y}=x^{\frac{3}{2}}+y^{\frac{3}{2}}=(x^\frac{1}{2})^3+(y^\frac{1}{2})^3=\\\\\star \; \; a^3+b^3=(a+b)(a^2-ab+b^2)\; \; \star \\\\=(x^{\frac{1}{2}}+y^{\frac{1}{2}})(x-x^{\frac{1}{2}}\cdot y^{\frac{1}{2}}+y)=(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y)$