a) ( x ² -1)( x +3) = 0 b) ( x ² +1)( x ² + 4x + 4) = 0 c) x ² -x -6 = 0 05/09/2021 Bởi Ruby a) ( x ² -1)( x +3) = 0 b) ( x ² +1)( x ² + 4x + 4) = 0 c) x ² -x -6 = 0
a) $( x ² -1)( x +3) = 0$ ⇔\(\left[ \begin{array}{}x²-1=0\\x+3=0\end{array} \right.\) ⇔\(\left[ \begin{array}{}x²=1\\x=-3\end{array} \right.\) ⇔\(\left[ \begin{array}{}x=±1\\x=-3\end{array} \right.\) b) $( x ² +1)( x ² + 4x + 4) = 0$ $⇔(x²+1)(x+2)²=0$ $⇔(x+2)²=0$ (vì $x²+1 >0$) $⇔x+2=0$ $⇔x=-2$ c) $x ² -x -6 = 0$ $⇔x²-3x+2x-6=0$ $⇔x(x-3)+2(x-3)=0$ $⇔(x-3)(x+2)=0$ ⇔\(\left[ \begin{array}{}x-3=0\\x+2=0\end{array} \right.\) ⇔\(\left[ \begin{array}{}x=3\\x=-2\end{array} \right.\) Bình luận
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a) $( x ² -1)( x +3) = 0$
⇔\(\left[ \begin{array}{}x²-1=0\\x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{}x²=1\\x=-3\end{array} \right.\)
⇔\(\left[ \begin{array}{}x=±1\\x=-3\end{array} \right.\)
b) $( x ² +1)( x ² + 4x + 4) = 0$
$⇔(x²+1)(x+2)²=0$
$⇔(x+2)²=0$ (vì $x²+1 >0$)
$⇔x+2=0$
$⇔x=-2$
c) $x ² -x -6 = 0$
$⇔x²-3x+2x-6=0$
$⇔x(x-3)+2(x-3)=0$
$⇔(x-3)(x+2)=0$
⇔\(\left[ \begin{array}{}x-3=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{}x=3\\x=-2\end{array} \right.\)