phân tích thành nhân tử a) (3x+1)^2 – 4(x-3)^2 =0 b) x^3+5x^2 + 4x +20 = 0 02/07/2021 Bởi Madeline phân tích thành nhân tử a) (3x+1)^2 – 4(x-3)^2 =0 b) x^3+5x^2 + 4x +20 = 0
a) `(3x+1)^2-4(x-3)^2=0` `⇒(3x+1)^2-(2x-6)^2=0` `⇒(3x+1-2x+6)(3x+1+2x-6)=0` `⇒5(x+7)(x-1)=0` $⇒\left[ \begin{array}{l}x+7=0\\x-1=0\end{array} \right.$ $⇒\left[ \begin{array}{l}x=-7\\x=1\end{array} \right.$ Vậy `x=-7; x=1` b) `x^3+5x^2+4x+20=0` `⇒x^2(x+5)+4(x+5)=0` `⇒(x+5)(x^2+4)=0` \(⇒\left[ \begin{array}{l}x+5=0\\x^2+4=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=-5\\x∈∅\end{array} \right.\) Vậy `x=-5` Bình luận
Đáp án: $a) (3x+1)^2 -4(x-3)^2 =0$ $⇔ [ 3x+1 -2(x-3) ] . [ 3x+1 +2(x-3) ] =0$ $⇔ (3x+1 -2x+6) . (3x+1 +2x-6)=0$ $⇔ (x+7)(5x-5)=0$ $b) x^3 +5x^2 +4x+20=0$ $⇔ x^2(x+5) +4(x+5)=0$ $⇔(x+5)(x^2+4)=0$ Bình luận
a) `(3x+1)^2-4(x-3)^2=0`
`⇒(3x+1)^2-(2x-6)^2=0`
`⇒(3x+1-2x+6)(3x+1+2x-6)=0`
`⇒5(x+7)(x-1)=0`
$⇒\left[ \begin{array}{l}x+7=0\\x-1=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=-7\\x=1\end{array} \right.$
Vậy `x=-7; x=1`
b) `x^3+5x^2+4x+20=0`
`⇒x^2(x+5)+4(x+5)=0`
`⇒(x+5)(x^2+4)=0`
\(⇒\left[ \begin{array}{l}x+5=0\\x^2+4=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=-5\\x∈∅\end{array} \right.\)
Vậy `x=-5`
Đáp án:
$a) (3x+1)^2 -4(x-3)^2 =0$
$⇔ [ 3x+1 -2(x-3) ] . [ 3x+1 +2(x-3) ] =0$
$⇔ (3x+1 -2x+6) . (3x+1 +2x-6)=0$
$⇔ (x+7)(5x-5)=0$
$b) x^3 +5x^2 +4x+20=0$
$⇔ x^2(x+5) +4(x+5)=0$
$⇔(x+5)(x^2+4)=0$