ĐKXĐ: \(x\ge-2\)
\(x^3-\left(y^3-3y^2+3y-1\right)+x-y+1=0\)
\(\Leftrightarrow x^3-\left(y-1\right)^3+x-y+1=0\)
\(\Leftrightarrow\left(x-y+1\right)\left(x^2+x\left(y-1\right)+\left(y-1\right)^2\right)+x-y+1=0\)
\(\Leftrightarrow\left(x-y+1\right)\left(x^2+x\left(y-1\right)+\left(y-1\right)^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y+1=0\\x^2+x\left(y-1\right)+\left(y-1\right)^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y+1=0\\x^2+x\left(y-1\right)+\dfrac{\left(y-1\right)^2}{4}+\dfrac{3\left(y-1\right)^2}{4}+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y+1=0\\\left(x+\dfrac{y-1}{2}\right)^2+\dfrac{3\left(y-1\right)^2}{4}+1=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x-y+1=0\Leftrightarrow y=x+1\) thay vào pt dưới:
\(x^3+x-3=2\sqrt{x+2}+x+1\)
\(\Leftrightarrow x^3-4=2\sqrt{x+2}\)
\(\Leftrightarrow x^3-8=2\left(\sqrt{x+2}-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=\dfrac{2\left(x-2\right)}{\sqrt{x+2}+2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Rightarrow x=2\Rightarrow y=3\\x^2+2x+4=\dfrac{2}{\sqrt{x+2}+2}\left(1\right)\end{matrix}\right.\)
Xét (1), ta có:
\(VT=\left(x+1\right)^2+3\ge3\) \(\forall x\)
\(\sqrt{x+2}\ge0\Rightarrow VP=\dfrac{2}{\sqrt{x+2}+2}\le\dfrac{2}{2}=1\) \(\forall x\ge-2\)
\(\Rightarrow VT>VP\Rightarrow\) pt (1) vô nghiệm
Vậy hệ đã cho có cặp nghiệm duy nhất \(\left(x;y\right)=\left(2;3\right)\)
