\(\left(a+10\right)+\left(a^2+10a\right)^2+2\left(a+10\right)^2=0\)
Có: \(\left\{{}\begin{matrix}\left(a^2+10a\right)^2\ge0\forall a\\2\left(a+10\right)^2\ge0\forall a\end{matrix}\right.\)
Để bt = 0 => \(\left\{{}\begin{matrix}\left(a^2+10a\right)^2=0\\2\left(a+10\right)^2=0\end{matrix}\right.\)\(\Rightarrow a=-10\)
Thay a = -10 vào a + 10 có: -10 + 10 = 0
(tm)
Vậy a = -10