Chọn D
\(\left(y-5\right)^{4} \) .Ta có: \(P=\sqrt{\left(x-2\right)^{4} +1} +\sqrt{\left(y-5\right)^{4} +1} \) Đặt \(\left\{\begin{array}{l} {u=x-2} \\ {v=y-5} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {x=u+2} \\ {y=v+5} \end{array}\right. . \)
Khi đó \(P=\sqrt{u^{4} +1} +\sqrt{v^{4} +1} \Rightarrow P\ge 2\)
Ta có \(xy-2x+y=27\Rightarrow \left(u+2\right)\left(v+5\right)-2\left(u+2\right)+v+5=27\Leftrightarrow uv+3u+3v=16.\)
Vì \(uv\le \frac{\left(u+v\right)^{2} }{4} \, \, \forall u,v \) nên \(16\le \frac{\left(u+v\right)^{2} }{4} +3\left(u+v\right)\Rightarrow \left[\begin{array}{l} {u+v\ge 4} \\ {u+v\le -16} \end{array}\right.\) .
\(P^{2} =a^{4} +b^{4} +2+2\sqrt{u^{4} v^{4} +u^{4} +v^{4} +1} \ge u^{4} +v^{4} +2+2\sqrt{u^{4} v^{4} +2u^{2} v^{2} +1} \) \(=u^{4} +v^{4} +2+2\left(u^{2} v^{2} +1\right)=\left(u^{2} +v^{2} \right)^{2} +4 \)
\( \Rightarrow P^{2} \ge \left(u^{2} +v^{2} \right)^{2} +4\ge \left(\frac{\left(u+v\right)^{2} }{2} \right)^{2} +4=\frac{\left(u+v\right)^{4} }{4} +4 \)
\(\Rightarrow P^{2} \ge \frac{4^{4} }{4} +4=68\Rightarrow P\ge 2\sqrt{17}\) ( do \(P\ge 2\)) \(\Rightarrow a=2;b=17\Rightarrow a+b=19.\)