Từ giả thiết đề bài ta có:$\\$ $\dfrac{x}{1 - x} + \dfrac{y}{1 - y} = 1$$\\$ $\Leftrightarrow \dfrac{x(1 - y) + y(1 - x)}{(1 - x)(1 - y)} = 1$$\\$ $\Leftrightarrow x - xy + y - xy = 1 - x - y + xy$$\\$ $\Leftrightarrow x + y - 2xy = 1 - (x + y) + xy$$\\$ $\Leftrightarrow 2(x + y) - 1 = 3xy$$\\$ $\Leftrightarrow 3xy = 2(x + y) - 1$$\\$ Lại có $\sqrt{x^2 - xy + y^2} = \sqrt{(x + y)^2 - 3xy}$$\\$ $= \sqrt{(x + y)^2 - [2(x + y) - 1]}$$\\$ $= \sqrt{(x + y)^2 - 2(x + y) + 1}$$\\$ $= \sqrt{(x + y - 1)^2}$$\\$ $= |x + y - 1|$$\\$ Vì $0 < x;y < 1$:$\\$ Ta có $\dfrac{x}{1 - x} > 0$ và $\dfrac{y}{1 - y} > 0$$\\$ Mà $\dfrac{x}{1 - x} + \dfrac{y}{1 - y} = 1 \Rightarrow \dfrac{x}{1 - x} < 1 \Rightarrow x < 1 - x \Rightarrow 2x < 1 \Rightarrow x < \dfrac{1}{2}$$\\$ Tương tự ta có $\dfrac{y}{1 - y} < 1 \Rightarrow y < 1 - y \Rightarrow 2y < 1 \Rightarrow y < \dfrac{1}{2}$$\\$ $\Rightarrow x + y < \dfrac{1}{2} + \dfrac{1}{2} = 1 \Rightarrow x + y - 1 < 0$$\\$ $\Rightarrow \vert x + y - 1 \vert = -(x + y - 1) = 1 - x - y$$\\$ Khi đó $P = x + y + 1 - x - y = 1$