Giải thích các bước giải:
a.Ta có:
ĐKXĐ:$ x\ge 0, y\ge 0, y\ne 1$
$P=\dfrac{x}{(\sqrt{x}+\sqrt{y})(1-\sqrt{y})}-\dfrac{y}{(\sqrt{x}+\sqrt{y})(\sqrt{x}+1)}-\dfrac{xy}{(\sqrt{x}+1)(1-\sqrt{y})}$
$\to P=\dfrac{x(\sqrt{x}+1)-y(1-\sqrt{y}) -xy(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{x\sqrt{x}+x-y+y\sqrt{y} -xy(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(x\sqrt{x}+y\sqrt{y}) +(x-y)-xy(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y) +(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})-xy(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}-xy)}{(\sqrt{x}+\sqrt{y})(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{x-\sqrt{xy}+y+\sqrt{x}-\sqrt{y}-xy}{(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(x+y-xy-1)+(\sqrt{x}-\sqrt{y}-\sqrt{xy}+1)}{(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(x+y-xy-1)+(\sqrt{x}-\sqrt{y}-\sqrt{xy}+1)}{(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(x-1)(1-y)+(1-\sqrt{y})(\sqrt{x}+1)}{(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)(1-\sqrt{y})(1+\sqrt{y})+(1-\sqrt{y})(\sqrt{x}+1)}{(1-\sqrt{y})(\sqrt{x}+1)}$
$\to P=\left(\sqrt{x}-1\right)\left(1+\sqrt{y}\right)+1$
b.Để $P=2$
$\to (\sqrt{x}-1)(\sqrt{y}+1)+1=2$
$\to (\sqrt{x}-1)(\sqrt{y}+1)=1$
$\to\sqrt{x}-1=\dfrac{1}{\sqrt{y}+1}\le\dfrac{1}{0+1}=1$
$\to \sqrt{x}\le 2$
$\to x\le 4$
Lại có $\sqrt{x}-1=\dfrac{1}{\sqrt{y}+1}>0$
$\to \sqrt{x}>1\to x>1$
Do $x\in Z\to x\in\{2,3,4\}$
$\to y\in\{2,\dfrac{2-\sqrt{3}}{2},0\}$
Vì $x,y\in Z\to (x,y)\in\{(2,2), (4,0)\}$
