Giải phương trình sau: -x^2 + 14x +120 = 0 07/08/2021 Bởi Josephine Giải phương trình sau: -x^2 + 14x +120 = 0
$-x²+14x+120=0$ $⇒x²-14x-120=0$ $⇒x²-20x+6x-120=0$ $⇒x(x-20)+6(x-20)=0$ $⇒(x-20(x+6)=0$ ⇒\(\left[ \begin{array}{l}x-20=0\\x+6=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=20\\x=-6\end{array} \right.\) $@nguyenduy28364$ Bình luận
`#DyHungg` `-x²+14x+120=0` `⇔-x²+20x-6x+120=0` `⇔-x(x-20)-6(x-20)=0` `⇔(x-20)(-x-6)=0` `1) x-20=0⇔x=20` `2) -x-6=0⇔x=-6` Vậy `S={-6;20}` Bình luận
$-x²+14x+120=0$
$⇒x²-14x-120=0$
$⇒x²-20x+6x-120=0$
$⇒x(x-20)+6(x-20)=0$
$⇒(x-20(x+6)=0$
⇒\(\left[ \begin{array}{l}x-20=0\\x+6=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=20\\x=-6\end{array} \right.\)
$@nguyenduy28364$
`#DyHungg`
`-x²+14x+120=0`
`⇔-x²+20x-6x+120=0`
`⇔-x(x-20)-6(x-20)=0`
`⇔(x-20)(-x-6)=0`
`1) x-20=0⇔x=20`
`2) -x-6=0⇔x=-6`
Vậy `S={-6;20}`