Question

Tìm x:  (x²-2)×(x+4)-4x(x-3)=-8+3x

Answer 1

\(\left(x^2-2\right)\cdot\left(x+4\right)-4x\left(x-3\right)=-8+3x\)

$\to$ \(x^3+4x^2-2x-8-4x^2+12x+8-3x=0\)

$\to$ \(x^3+7x=0\)

$\to x.(x^2+7)=0$

$\to x=0$ ( Do $x^2+7>0$ )

Vậy $x=0$

Answer 2

\(\Leftrightarrow x^3-2x+4x^2-8-4x^2+12x+8-3x=0\)

\(\Leftrightarrow x^3+7x=0\)

\(\Leftrightarrow x\left(x^2+7\right)=0\)

Mà \(x^2+7\ge7>0\)

\(\Leftrightarrow x=0\)

Vậy ...

Answer 3

`(x^2-2)(x+4)-4x(x-3)=-8+3x`

`<=>x^3+4x^2-2x-8-4x^2+12x-3x+8=0`

`<=>x^3+7x=0`

`<=>x(x^2+7)=0`

Vì `x^2+7>=7>0`

`<=>x=0`