Đáp án: $ P=\sqrt x+\sqrt{xy}-\sqrt{y}$

Giải thích các bước giải:

a.ĐKXĐ: $x\ge 0, y\ge 0, y\ne 1, x^2+y^2\ne 0$

Ta có:

$P=\dfrac{x}{(\sqrt x+\sqrt y)(1-\sqrt y)}-\dfrac{y}{(\sqrt x+\sqrt y)(\sqrt x+1)}-\dfrac{xy}{(\sqrt x+1)(1-\sqrt y)}$

$\to P=\dfrac{x(1+\sqrt x)-y(1-\sqrt y)-xy(\sqrt x+\sqrt y)}{(\sqrt x+\sqrt y)(1-\sqrt y)(1+\sqrt x)}$

$\to P=\dfrac{ (x-y)+(x\sqrt{x}+y\sqrt{y})-xy(\sqrt{x}+\sqrt{y})}{(\sqrt x+\sqrt y)(1-\sqrt y)(1+\sqrt x)}$

$\to P=\dfrac{ (\sqrt x-\sqrt y)(\sqrt x+\sqrt y)+(\sqrt x+\sqrt y)(x-\sqrt{xy}+y)-xy(\sqrt{x}+\sqrt{y})}{(\sqrt x+\sqrt y)(1-\sqrt y)(1+\sqrt x)}$

$\to P=\dfrac{ (\sqrt x-\sqrt y)+(x-\sqrt{xy}+y)-xy}{(1-\sqrt y)(1+\sqrt x)}$

$\to P=\dfrac{\sqrt{x}(\sqrt{x}+1)-\sqrt{y}(\sqrt{x}+1)+y(1-\sqrt x)(1+\sqrt x)}{(1-\sqrt y)(1+\sqrt x)}$

$\to P=\dfrac{\sqrt{x}-\sqrt{y}+y(1-\sqrt x)}{1-\sqrt y}$

$\to P=\dfrac{\sqrt{x}-\sqrt{y}+y-y\sqrt{x}}{1-\sqrt y}$

$\to P=\dfrac{\sqrt{x}(1-\sqrt y)(1+\sqrt y)-\sqrt y(1-\sqrt y)}{1-\sqrt y}$

$\to P=\sqrt{x}(1+\sqrt y)-\sqrt y$

$\to P=\sqrt x+\sqrt{xy}-\sqrt{y}$