Giải thích các bước giải:
$C=-9x^2+6x-2$
$=-(9x^2-6x+2)$
$=-[(3x)^2-2.3x.1+1^2+1]$
$=-(3x-1)^2-1$
Ta có:
$-(3x-1)^2\le 0$
$⇒-(3x-1)^2-1<0$
$⇒$ Đa thức vô nghiệm.
Đáp án + Giải thích các bước giải:
`C=0`
`=>-9x^2+6x-2=0`
`=>-(9x^2-6x+1)-1=0`
`=>-(3x-1)^2-1=0`
Với mọi `x` có: `(3x-1)^2\ge0`
`=>-(3x-1)^2\le0`
`=>-(3x-1)^2-1\le-1<0`
Vậy đa thức vô nghiệm
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