Đáp án:
$Min_{3x^2-x+2}=\frac{23}{12}$ `⇔x=\frac{1}{6}`
Giải thích các bước giải:
Ta có :
`3x^2-x+2`
`=3(x^2-\frac{x}{3}+\frac{2}{3})`
`=3[x^2-2.x.\frac{1}{6}+(\frac{1}{6})^2-(\frac{1}{6})^2+\frac{2}{3}]`
`=3[x^2-2.x.\frac{1}{6}+(\frac{1}{6})^2+\frac{23}{36}]`
`=3[x^2-2.x.\frac{1}{6}+(\frac{1}{6})^2]+\frac{23}{12}`
`=3(x-\frac{1}{6})^2+\frac{23}{12}≥\frac{23}{12}`
Dấu '' = '' xảy ra khi :
`(x-\frac{1}{6})^2=0`
`⇔x=\frac{1}{6}`
Vậy $Min_{3x^2-x+2}=\frac{23}{12}$ `⇔x=\frac{1}{6}`
