Toán Lớp 8: tìm x
x(x^2-4)=0
(x+2)^2-(x-2)(x+2)=0
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Toán Lớp 8: tìm x
x(x^2-4)=0
(x+2)^2-(x-2)(x+2)=0
Comments ( 2 )
$x(x^2-4)=0$
$⇒$ \(\left[ \begin{array}{l}x=0\\x^2-4=0\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=0\\x^2=4\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\)
Vậy $x=0$ hoặc $x=±2$
$b,$ $(x+2)^2-(x-2)(x+2)=0$
$(x+2)(x+2-x+2)=0$
$(x+2).4=0$
$x+2=0$
$⇒x=-2$
Vậy $x=-2$
x(x^2-4)=0
$⇔\left[\begin{matrix} x=0\\ x^2-4=0\end{matrix}\right.$
$⇔\left[\begin{matrix} x=0\\ x^2=4\end{matrix}\right.$
$⇔\left[\begin{matrix} x=0\\ x=±2\end{matrix}\right.$
Vậy x in {0;2;-2}
(x+2)^2-(x-2)(x+2)=0
<=>x^2+4x+4-(x^2-4)=0
<=>x^2+4x+4-x^2+4=0
<=>4x+8=0
<=>4x=-8
<=>x=-2
Vậy x=-2