Bài 1:
Áp dụng BĐT: $a^2+b^2\le(a+b)^2\le2(a^2+b^2),a,b\ge0$ ta có:
$(x+y)^2=9(\sqrt{x+1}+\sqrt{y+2})^2\le 18(x+y+3) \Rightarrow x+y\le 9+3\sqrt{15}$
$\max P=9+3\sqrt{15} \Leftrightarrow \left\{\begin{array}{l}x=\dfrac{10+3\sqrt{15}}{2}\\y=\dfrac{8+3\sqrt{15}}{2}\end{array}\right.$
$(x+y)^2=9(\sqrt{x+1}+\sqrt{y+2})^2\ge 9(x+y+3) \Rightarrow x+y\ge \dfrac{9+3\sqrt{21}}{2}$
$\min P=\dfrac{9+3\sqrt{21}}{2}\Leftrightarrow \left[\begin{array}{l}x=-1;y=\dfrac{11+3\sqrt{21}}{2}\\x=\dfrac{13+3\sqrt{21}}{2};y=-2\end{array}\right.$