Question

Answer 1

Với `x >= 0,y >= 0,x \ne y` có:

`[x-2\sqrt{xy}+y]/[x-y]+[2\sqrt{y}]/[\sqrt{x}+\sqrt{y}]`

`=[(\sqrt{x}-\sqrt{y})^2]/[(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})]+[2\sqrt{y}]/[\sqrt{x}+\sqrt{y}]`

`=[\sqrt{x}-\sqrt{y}]/[\sqrt{x}+\sqrt{y}+[2\sqrt{y}]/[\sqrt{x}+\sqrt{y}]`

`=[\sqrt{x}+\sqrt{y}]/[\sqrt{x}+\sqrt{y}]=1`

Answer 2

\(21,\dfrac{x-2\sqrt{xy}+y}{x-y}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x-2\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{2\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\dfrac{x-2\sqrt{xy}+y+2\sqrt{xy}-2y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\dfrac{x-y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =1\)