Đáp án:

$P_{\min}= \sqrt5- 1$

Giải thích các bước giải:

\(\begin{array}{l} \quad \log_2\dfrac{1-xy}{x+y} = 2xy + x + y - 3\qquad (x;y>0; xy < 1)\\ \Leftrightarrow \log_2(1 - xy) + 3 -2xy = \log_2(x+y) + x + y\\ \Leftrightarrow \log_22(1-xy) + 2(1-xy) = \log_2(x+y) + x + y\qquad (*)\\ \text{Xét}\ f(t) = \log_2t + t,\ \ t>0\\ \Rightarrow f'(t) = \dfrac{1}{t\ln2} + 1 >0\quad \forall t >0\\ \Rightarrow f(t)\ \text{đồng biến trên $(0;+\infty)$}\\ \text{Do đó:}\\ (*)\Leftrightarrow 2(1 - xy) = x + y\\ \Leftrightarrow x  +2xy = 2 - y\\ \Leftrightarrow x = \dfrac{2 - y}{1 + 2y}\\ \text{Do}\ \begin{cases}x >0\\1 + 2y >0\end{cases}\\ \text{nên}\ 2 -y >0\Leftrightarrow 0 <y < 2\\ \text{Ta được:}\\ \quad P = \dfrac{2-y}{1+2y} + y\\ \text{Đặt}\ P = f(y) = \dfrac{2-y}{1+2y} + y\\ \Rightarrow f'(y) = \dfrac{4(y^2+ y - 1)}{(1+2y)^2},\ \ \forall y\in \left(0;2\right)\\ f'(y) = 0 \Leftrightarrow \left[\begin{array}y =\dfrac{-1-\sqrt5}{2}\\y = \dfrac{-1+\sqrt5}{2}\end{array}\right.\\ \text{Bảng biến thiên $f(y)$ trên $\left(0;2\right):$}\\ \begin{array}{c|ccc} y&0&&\dfrac{-1+\sqrt5}{2}&&2\\\hline f'(y)&\vert&-&0&+&\vert\\\hline &\\ f(y)&&\searrow&&\nearrow\\ &&&\sqrt5-1 \end{array}\\ \text{Dựa vào bảng biến thiên, ta được:}\\ \mathop{\min}\limits_{\left(0;2\right)}f(y) = \sqrt5-1\Leftrightarrow y = \dfrac{-1 + \sqrt{5}}{2} \Rightarrow x = \dfrac{-1 + \sqrt{5}}{2}\\ \text{Vậy}\ P_{\min} =\sqrt5-1\Leftrightarrow (x;y) = \left(\dfrac{-1 + \sqrt{5}}{2};\dfrac{-1 + \sqrt{5}}{2}\right) \end{array}\)