`(x + 1)^2 + (x^2 - 1)^2=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=-1\end{matrix}\right.\)
\(x^2+2x+1+x^4-2x^2+1=0\\ x^4-x^2+2x+2=0\\ x^2\left(x^2-1\right)+2\left(x+1\right)=0\\ \left(x+1\right)\left[x^2\left(x-1\right)+2\right]=0\\ \left[{}\begin{matrix}x+1=0\\x^3-x^2+2=0\end{matrix}\right.=>x=-1}\)
\(\left(x+1\right)^2+\left(x^2-1\right)^2=0\)
\(\Leftrightarrow x^2+2x+1+x^4-2x^2+1=0\)
\(\Leftrightarrow x^4-x^2+2x+2=0\)
\(\Leftrightarrow x^2.\left(x^2-1\right)+2.\left(x+1\right)=0\)
\(\Leftrightarrow x^2.\left(x+1\right).\left(x-1\right)+2.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left[x^2.\left(x-1\right)+2\right]=0\)
\(\Leftrightarrow\left(x+1\right).\left(x^3-x^2+2\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x^3+x^2-2x^2-2x+2x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left[x^2.\left(x+1\right)-2x.\left(x+1\right)+2.\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right).\left(x^2-2x+2\right).\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\) hoặc \(\Leftrightarrow x^2-2x+2=0\)
\(\Leftrightarrow x=-1\) \(\Leftrightarrow x^2-2x+1+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=-1\)(vô lí)
Vậy phương trình có tập nghiệm \(S=\left\{-1\right\}\)