2$x^{3}$ + 7$x^{2}$ +4x-4=0 tìm x

Đáp án:

$\text{x $\in$ {-2 ; $\dfrac{1}{2}$}}$ 

Giải thích các bước giải:

Ta có:

$2x^3+7x^2+4x-4=0$

$⇔\left(x+2\right)^2\left(2x-1\right)=0$

$⇔x+2=0\quad \mathrm{hoặc}\quad \:2x-1=0$

TH1: $x+2=0$

$⇔x+2-2=0-2$

$⇔x=-2$

`------`

TH2: $2x-1=0$

$⇔2x-1+1=0+1$

$⇔2x=1$

$⇔\dfrac{2x}{2}=\dfrac{1}{2}$

$⇔x=\dfrac{1}{2}$

$\text{Vậy x $\in$ {-2 ; $\dfrac{1}{2}$}}$

$2x^3+7x^2+4x-4=0$

$⇒\left(x+2\right)^2\left(2x-1\right)=0$

$⇒x+2=0\quad \mathrm{hoặc}\quad \:2x-1=0$

TH1: $x+2=0$

$⇒x+2-2=0-2$

$⇒x=-2$

TH2: $2x-1=0$

$⇒2x-1+1=0+1$

$⇒2x=1$

$⇒\dfrac{2x}{2}=\dfrac{1}{2}$

$⇒x=\dfrac{1}{2}$

$\text{Vậy x $\in$ {-2 ; $\dfrac{1}{2}$}}$