ĐKXĐ: \(\left\{{}\begin{matrix}x\le1\\x+2y\ge1\end{matrix}\right.\) và \(y\ge0\)
\(2y^3-2y^2=x^2+3xy-xy^2\)
\(\Leftrightarrow2y^3+xy^2-\left(x^2+3xy+2y^2\right)=0\)
\(\Leftrightarrow y^2\left(x+2y\right)-\left(x+y\right)\left(x+2y\right)=0\)
\(\Leftrightarrow\left(y^2-x-y\right)\left(x+2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2y=0\left(l\right)\\x=y^2-y\end{matrix}\right.\)
\(\Rightarrow\sqrt{y^2+y-1}+\sqrt{1+y-y^2}=y^2-y+2\)
Mặt khác ta có:
\(VT=\sqrt{y^2+y-1}+\sqrt{1+y-y^2}\le\sqrt{2.2y}=2\sqrt{y}\)
\(\Rightarrow y^2-y+2\le2\sqrt{y}\)
\(\Leftrightarrow y\left(y-1\right)-2\left(\sqrt{y}-1\right)\le0\)
\(\Leftrightarrow\left(\sqrt{y}-1\right)^2\left(y+\sqrt{y}+2\right)\le0\)
\(\Leftrightarrow\sqrt{y}=1\Rightarrow y=1\Rightarrow x=0\)
Vậy hệ có nghiệm duy nhất \(\left(x;y\right)=\left(0;1\right)\)
