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