Đáp án:
Giải thích các bước giải:
1 $(2,3x+6,9)(4-x²) = 0$
⇔\(\left[ \begin{array}{l}2,3x+6,9=0\\4-x²=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-6,9÷2,3\\-x²=-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x²=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=±2\end{array} \right.\)
3 $- x² + x + 6 = 0$
$⇔ -x² - 2x + 3x +6 = 0$
$⇔ -x(x+2) + 3(x+2) = 0$
$⇔ (x+2)(-x+3)=0$
⇔\(\left[ \begin{array}{l}x+2=0\\-x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)