bài 1: tìm x C) (x+5) *(x^2 -4)=0 D) (x+3)*(x^2 +1)=0 17/08/2021 Bởi Isabelle bài 1: tìm x C) (x+5) *(x^2 -4)=0 D) (x+3)*(x^2 +1)=0
c) $(x+5)×(x²-4)=0$ ⇒ \(\left[ \begin{array}{l}x+5=0\\x²-4=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=-5\\x²=4\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=-5\\x=±2\end{array} \right.\) d) $(x+3)×(x²+1)=0$ ⇒ \(\left[ \begin{array}{l}x+3=0\\x²+1=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=-3\\x²=-1(loại)\end{array} \right.\) ⇒ $x=-3$ Bình luận
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 Bình luận
c) $(x+5)×(x²-4)=0$
⇒ \(\left[ \begin{array}{l}x+5=0\\x²-4=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-5\\x²=4\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-5\\x=±2\end{array} \right.\)
d) $(x+3)×(x²+1)=0$
⇒ \(\left[ \begin{array}{l}x+3=0\\x²+1=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-3\\x²=-1(loại)\end{array} \right.\)
⇒ $x=-3$
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