$\frac{x-3}{-2}$ $=$ $\frac{-8}{x-3}$ ⇒ (x-3)(x-3) = (-8)(-2) ⇔ x² – 6x + 9 = 16 ⇔ x² -6x – 7 = 0 ⇔ x² + x -7x -7 = 0 ⇔ x(x+1) – 7(x+1) = 0 ⇔(x-7)(x+1) = 0 ⇔\(\left[ \begin{array}{l}x-7=0\\x+1=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\) Trả lời
$\dfrac{x-3}{-2}=\dfrac{-8}{x-3}$ $⇒(x-3).(x-3)=(-2).(-8)$ $⇒(x-3)²=16$ $⇒(x-3)²=(±4)²$ $⇒x-3=±4$ $⇒$ \(\left[ \begin{array}{l}x-3=4\\x-3=-4\end{array} \right.\) $⇒$ \(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\) Trả lời
$\frac{x-3}{-2}$ $=$ $\frac{-8}{x-3}$
⇒ (x-3)(x-3) = (-8)(-2)
⇔ x² – 6x + 9 = 16
⇔ x² -6x – 7 = 0
⇔ x² + x -7x -7 = 0
⇔ x(x+1) – 7(x+1) = 0
⇔(x-7)(x+1) = 0
⇔\(\left[ \begin{array}{l}x-7=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\)
$\dfrac{x-3}{-2}=\dfrac{-8}{x-3}$
$⇒(x-3).(x-3)=(-2).(-8)$
$⇒(x-3)²=16$
$⇒(x-3)²=(±4)²$
$⇒x-3=±4$
$⇒$ \(\left[ \begin{array}{l}x-3=4\\x-3=-4\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\)