Chọn D
Cách 1.
Ta có: \(\left\{\begin{array}{l} {1\le x\le 2} \\ {1\le y\le 2} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {\left(x-1\right)\left(x-2\right)\le 0} \\ {\left(y-1\right)\left(y-2\right)\le 0} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {x^{2} \le 3x-2} \\ {y^{2} \le 3y-2} \end{array}\right. .\)
Khi đó: \(P\ge \frac{3\left(x+y\right)-5\left(x+y\right)+2xy}{x+y-1} +4\left(\frac{x+2y}{3x+3y-3} \right)+4\left(\frac{y+2x}{3y+3x-3} \right).\)
\(\[\Rightarrow P\ge \frac{x^{2} +y^{2} +2xy-5\left(x+y\right)+4}{x+y-1} +\frac{4\left(x+y\right)}{x+y-1} .\] \)
\(\[\Rightarrow P\ge \frac{\left(x+y\right)^{2} -\left(x+y\right)+4}{x+y-1} .\] \)
Đặt \( t=x+y vì 2\le x+y\le 4 nên t\in \left[2;\, 4\right].\)
Xét hàm số \(f\left(t\right)=\frac{t^{2} -t+4}{t-1} với \forall t\in \left[2;\, 4\right].\)
Ta có \(f'\left(t\right)=\frac{t^{2} -2t-3}{\left(t-1\right)^{2} } , f'\left(t\right)=0\Leftrightarrow \left[\begin{array}{l} {t=3} \\ {t=-1} \end{array}\right. .\)
\(\[f\left(2\right)=6,\, f\left(3\right)=5,\, f\left(4\right)=\frac{16}{3} .\] \)
Vì hàm số liên tục trên \(\left[2;\, 4\right] nên \min P={\mathop{\min }\limits_{\left[2;\, 4\right]}} f\left(t\right)=5 khi \left(x;\, \, y\right)\in \left\{\left(1;\, 2\right),\, \left(2;\, 1\right)\right\}.\)
Cách 2.
Ta có: \(\left\{\begin{array}{l} {1\le x\le 2} \\ {1\le y\le 2} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {\left(x-1\right)\left(x-2\right)\le 0} \\ {\left(y-1\right)\left(y-2\right)\le 0} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {x^{2} \le 3x-2} \\ {y^{2} \le 3y-2} \end{array}\right. .\)
Khi đó: \( P\ge 2\left[\frac{xy-\left(x+y\right)}{x+y-1} \right]+4\left(\frac{x+2y}{3x+3y-3} \right)+4\left(\frac{y+2x}{3y+3x-3} \right).\)
\(\[P\ge \frac{2\left[xy-\left(x+y\right)\right]+4\left(x+y\right)}{x+y-1} =2\left(\frac{xy+x+y}{x+y-1} \right)=2\left(\frac{xy+1}{x+y-1} +1\right).\] \)
Ta chứng minh: \(\frac{xy+1}{x+y-1} \ge \frac{3}{2} với \forall x,y\in \left[1;\, 2\right].\)
Thật vậy:\( \frac{xy+1}{x+y-1} \ge \frac{3}{2} \Leftrightarrow 2xy-3\left(x+y\right)+5\ge 0\Leftrightarrow \left(x-1\right)\left(y-1\right)+\left(x-2\right)\left(y-2\right)\ge 0 luôn đúng với \forall x,y\in \left[1;\, 2\right].\)
Do đó: \(P\ge 2\left(\frac{3}{2} +1\right)=5.\)
Dấu ''='' xảy ra khi \(\left(x;\, \, y\right)\in \left\{\left(1;\, 2\right),\, \left(2;\, 1\right)\right\}.\)
Vậy\( \min P=5.\)