Đặt \(\sqrt{1-x}=z\ge0\Rightarrow x=1-z^2\)
\(\Rightarrow2y^3+7y+2\left(1-z^2\right)z=3z+6y^2+3\)
\(\Leftrightarrow2y^3+7y+2z-2z^3=3z+6y^2+3\)
\(\Leftrightarrow2y^3-6y^2+7y-3=2z^3+z\)
\(\Leftrightarrow2\left(y-1\right)^3+\left(y-1\right)=2z^3+z\)
Hàm \(f\left(t\right)=2t^3+t\) có \(f'\left(t\right)=6t^2+1>0;\forall t\) nên đồng biến trên R
\(\Rightarrow y-1=z\)
\(\Rightarrow y=1+\sqrt{1-x}\)
\(P=x+2\left(1+\sqrt{1-x}\right)=x+2\sqrt{1-x}+2\)
\(=-\left(1-x-2\sqrt{1-x}+1\right)+4=-\left(\sqrt{1-x}-1\right)^2+4\le4\)