Question

2x³ − 3x² + x + 6 = 0

 

Answer 1

`2x^3-3x^2+x+6=0`

`<=>2x^3+2x^2-5x^2-5x+6x+6=0`

`<=>2x^2(x+1)-5x(x+1)+6(x+1)=0`

`<=>(x+1)(2x^2+5x+6)=0`

Vì `2x^2+5x+6`

`=2(x+5/4)^2+23/8>=23/8>0`

`=>x+1=0<=>x=-1`

vậy `S={-1}`

Answer 2

Ta có : \(2x^3-3x^2+x+6=0\)

\(\Leftrightarrow2x^3+2x^2-5x^2-5x+6x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+6\right)=0\)

Thấy : \(2x^2-5x+6\ge\dfrac{23}{8}>0\)

=> x + 1 = 0

<=> x = -1 

Vậy ...

Answer 3

\(2x^3-3x^2+x+6=0\Leftrightarrow2x^3+2x^2-5x^2-5x+6x+6=0\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+6\left(x+1\right)=0\)\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+6\right)=0\)

Ta thấy: \(2x^2-5x+6=2\left(x^2-\dfrac{5}{2}x+3\right)=2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}+\dfrac{23}{16}\right)=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{2.23}{16}=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{23}{8}\)\(\ge\dfrac{23}{8}>0\ne0\)

=> x+1=0 <=>x=-1

Answer 4

Ta có: \(2x^3-3x^2+x+6=0\)

\(\Leftrightarrow2x^3+2x^2-5x^2-5x+6x+6=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+6\right)=0\)

mà \(2x^2-5x+6>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}