Đk: x >/ 0
(1) \(\Leftrightarrow x\left(x-y\right)+y^2\left(x-y\right)=0\Leftrightarrow\left(x+y^2\right)\left(x-y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y^2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y^2\left(vô-lý,x\ge0\right)\\x=y\end{matrix}\right.\)
Với x=y, thay vào (2), ta được:
\(2\left(x^2+1\right)-3\sqrt{x}\left(x+1\right)-x=0\)
\(\Leftrightarrow2x^2+2-3x\sqrt{x}-3\sqrt{x}-x=0\) (*)
Đặt \(t=\sqrt{x}\left(t\ge0\right)\) , pt (*) trở thành:
\(2t^4+2-3t^3-3t-t^2=0\Leftrightarrow\left(t-2\right)\left(2t^3+t^2+t-1\right)=0\Leftrightarrow\left(t-2\right)\left(2t-1\right)\left(2t^2+2t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=\dfrac{1}{2}\\2t^2+2t+2=0\left(vn\right)\end{matrix}\right.\)
(+) Với t=2, ta có: \(\sqrt{x}=2\Leftrightarrow x=4\left(n\right)\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\left(N\right)\)
(+) Với t=1/2, ta có: \(\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(n\right)\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\end{matrix}\right.\left(N\right)\)
Kl: \(\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\), \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\end{matrix}\right.\)
