Giải thích các bước giải:
$2y^3+7y+2x\sqrt{1-x}=3\sqrt{1-x}+3(2y^2+1)$
$\to 2(y^3-3y^2+3y-1)+y+2(x-1)\sqrt{1-x}=\sqrt{1-x}+1$
$\to 2(y-1)^3+(y-1)=2(1-x)\sqrt{1-x}-\sqrt{1-x}$
$\to y-1=\sqrt{1-x}$
$\to y=1+\sqrt{1-x}$
$\to x+2y=x+2+2\sqrt{1-x}=4-(\sqrt{1-x}^2-2\sqrt{1-x}+1)=4-(\sqrt{1-x}-1)^2\le 4$
$\to P=4$
