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Đặt: $t=x+y;t>2$
Ta có: $(x+y)^2\ge4xy\Rightarrow xy\le\dfrac{t^2}{4}$
Ta có:
$P=\dfrac{(x+y)^3-3xy(x+y)-(x+y)^2+2xy}{xy-(x+y)+1}$
$=\dfrac{t^3-t^2-xy(3t-2)}{xy-t+1}$
$=\dfrac{t^3-t^2-\dfrac{t^2(3t-2)}{4}}{\dfrac{t^2}{4}-t+1}=\dfrac{t^2}{t-2}$
Xét hàm: $f(t)=\dfrac{t^2}{t-2};t>2$
Ta có: $f'(t)=\dfrac{t^2-4t}{(t-2)^2}$
$f'(t)=0 \Leftrightarrow \left[\begin{array}{l}t=0\\t=4\end{array}\right. \Leftrightarrow t=4$
Lập bảng biến thiên ta có: $\displaystyle \min_{t\in(2;+\infty)}f(t)=f(4)=8$
Vậy: Min$P=8 \Leftrightarrow \left\{\begin{array}{l}x+y=4\\xy=4\end{array}\right. \Leftrightarrow x=y=2$
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