a) \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)
\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)
b) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+x+5\right)=0\)
\(\Rightarrow x+1=0\left(2x^2+x+5\ne0\forall x\right)\)
<=> x=-1
Vậy x=-1
