ĐKXĐ:$x;y \geq 0;x \neq y$
$A=[\dfrac{(\sqrt[]x-\sqrt[]y)(\sqrt[]x+\sqrt[]y)}{\sqrt[]x-\sqrt[]y}-\dfrac{(\sqrt[]x-\sqrt[]y)(x+\sqrt[]{xy}+y)}{(\sqrt[]x-\sqrt[]y)(\sqrt[]x+\sqrt[]y)}]:[\dfrac{x-\sqrt[]{xy}+y}{\sqrt[]x+\sqrt[]y}]$
$=[\sqrt[]x+\sqrt[]y-\dfrac{x+\sqrt[]{xy}+y}{\sqrt[]x+\sqrt[]y}].\dfrac{\sqrt[]x+\sqrt[]y}{x-\sqrt[]{xy}+y}$
$=(\sqrt[]x+\sqrt[]y).\dfrac{\sqrt[]x+\sqrt[]y}{x-\sqrt[]{xy}+y}-1$
$=\dfrac{(\sqrt[]x+\sqrt[]y)^2}{x-\sqrt[]{xy}+y}-1$
$=\dfrac{x+2.\sqrt[]{xy}+y-x+\sqrt[]{xy}-y}{x-\sqrt[]{xy}+y}$
$=\dfrac{3.\sqrt[]{xy}}{x-\sqrt[]{xy}+y}$
b, Ta có:
$A=\dfrac{(\sqrt[]x+\sqrt[]y)^2}{x-\sqrt[]{xy}+y}-1$
(Dòng thứ 5 nhé bạn)
Mà $(\sqrt[]x+\sqrt[]y)^2≥0;x-\sqrt[]{xy}+y=(\sqrt[]x-\dfrac{1}{2})^2+\dfrac{3}{4}>0$
$⇒\dfrac{(\sqrt[]x+\sqrt[]y)^2}{x-\sqrt[]{xy}+y}≥0$
$⇒A=\dfrac{(\sqrt[]x+\sqrt[]y)^2}{x-\sqrt[]{xy}+y}-1≥-1$
Dấu = xảy ra $⇔\sqrt[]x+\sqrt[]y=0⇔\sqrt[]x=0;\sqrt[]y=0⇔x=y=0$
