\(\hept{\begin{cases}y^6+y^3+2x^2=\sqrt{xy-x^2y^2}\left(1\right)\\4xy^3+y^2+\frac{1}{2}\ge2x^2+\sqrt{1+\left(2x-y\right)^2}\left(2\right)\end{cases}}\)
\(VP\left(1\right)=\sqrt{\frac{1}{4}-\left(xy-\frac{1}{2}\right)^2}\le\frac{1}{2}\Rightarrow VT\left(1\right)=y^6+y^3+2x^2\le\frac{1}{2}\)
\(\Leftrightarrow2x^2+2y^3+4x^2\le1\left(3\right)\)
Từ (2)(3) => \(8xy^3+2y^3+2\ge2y^6+4x^2+4x^2+2\sqrt{1+\left(2x-y\right)^2}\)
\(\Leftrightarrow8xy^3+2\ge2y^6+8x^2+2\sqrt{2+\left(2x-y\right)^2}\)
\(\Leftrightarrow4xy^3+1\ge y^6+4x^2+\sqrt{1+\left(2x-y\right)^2}\)
\(\Leftrightarrow1-\sqrt{1+\left(2x-y\right)^2}\ge y^6-4xy^3+4x^2=\left(y^3-2x\right)^2\left(4\right)\)
\(VT\left(4\right)\le0;VP\left(4\right)\ge0\). Do đó:
(4) \(\Leftrightarrow\hept{\begin{cases}y=2x\\y^3=2x\end{cases}\Leftrightarrow\hept{\begin{cases}y=2x\\y^3=y\end{cases}}}\)<=> \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)hoặc \(\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}\)hoặc \(\hept{\begin{cases}x=\frac{-1}{2}\\y=-1\end{cases}}\)
Thử lại chỉ có \(\left(x;y\right)=\left(\frac{-1}{2};-1\right)\)thỏa mãn
Vậy hệ đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(\frac{-1}{2};-1\right)\)
