Question

a) 3x³ + 10x² − 8x = 0

b) (x² + 5)(x - 2) + 4 = 2x

Answer 1

`a)3x^3+10x^2-8x=0`

`<=>x(3x^2+10x-8)=0`

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

`<=>x(x+4)(3x-2)=0`

`<=>[(x=0),(x=-4),(x=2/3):}`

Vậy `S={0;-4;2/3}`

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`b)(x^2+5)(x-2)+4=2x`

`<=>(x^2+5)(x-2)-(2x-4)=0`

`<=>(x^2+5)(x-2)-2(x-2)=0`

`<=>(x-2)(x^2+5-2)=0`

`<=>(x-2)(x^2+3)=0`

 Mà `x^2+3 > 0`

  `=>x-2=0`

`<=>x=2`

Vậy `S={2}`

Answer 2

a) \(3x^3+10x^2-8x=0\)

\(\Rightarrow x\left(3x^2+10x-8\right)=0\\ \Rightarrow x\left(x+4\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

b) \(\left(x^2+5\right)\left(x-2\right)+4=2x\)

\(\Rightarrow\left(x^2+5\right)\left(x-2\right)+4-2x=0\)

\(\Rightarrow\left(x^2+5\right)\left(x-2\right)-2\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x^2+3\right)=0\\ \Rightarrow x-2=0\\ \Rightarrow x=2\)

(Do \(x^2+3>0,\forall x\))

Answer 3

a)3x3+10x2−8x=0a)3x3+10x2-8x=0

⇔x(3x2+10x−8)=0⇔x(3x2+10x-8)=0

⇔x(3x2+12x−2x−8)=0⇔x(3x2+12x-2x-8)=0

⇔x(x+4)(3x−2)=0⇔x(x+4)(3x-2)=0

⇔⎡⎢ ⎢⎣x=0x=−4x=23⇔[x=0x=-4x=23

Vậy S={0;−4;23}