a) `(3x+1)^2-4(x-3)^2=0`
`⇒(3x+1)^2-(2x-6)^2=0`
`⇒(3x+1-2x+6)(3x+1+2x-6)=0`
`⇒5(x+7)(x-1)=0`
$⇒\left[ \begin{array}{l}x+7=0\\x-1=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=-7\\x=1\end{array} \right.$
Vậy `x=-7; x=1`
b) `x^3+5x^2+4x+20=0`
`⇒x^2(x+5)+4(x+5)=0`
`⇒(x+5)(x^2+4)=0`
\(⇒\left[ \begin{array}{l}x+5=0\\x^2+4=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=-5\\x∈∅\end{array} \right.\)
Vậy `x=-5`