Bài 1 :
C, ( x + 5 ) . ( x² - 4 ) = 0
⇒ \(\left[ \begin{array}{l}x+5=0\\x²-4=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-5\\x²=4\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-5\\x∈{-2 ; 2 }\end{array} \right.\)
Vậy x ∈ { ±2 ; 5 }
D, ( x + 3 ) . ( x² + 1 ) = 0
⇒ \(\left[ \begin{array}{l}x+3=0\\x²+1=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-3\\x²=-1\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-3\\x∈∅\end{array} \right.\)
Vậy x = -3
