có nghiệm vì
\(\left(x-2\right)^2>=0\)
\(\left(x+1\right)^2>=0\)
\(=>\left(x-2\right)^2+\left(x+1\right)^2có\)\(nghiệm\)
\(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x+1\right)^2\ge0\end{cases}}\forall x\Rightarrow\left(x-2\right)^2+\left(x+1\right)^2\ge0\forall x\)
Vậy...