Đáp án:
`Q=(3x^2+9x+17)/(3x^2+9x+7)=(3x^2+9x+7+10)/(3x^2+9x+7)`
`=1+10/(3x^2+9x+7)`
Ta có: `3x^2+9x+7`
`=3.(x^2+3x+7/3)`
`=3.[x^2+2.x. 9/2+(9/2)^2-(9/2)^2+7/3]`
`=3.[(x+9/2)^2-215/72]>=-215/72`
`=3(x+9/2)^2-215/4`
`=> 10/(3x^2+9x+7)<=10/(-215/4)=-8/43`
`=> 1+10/(3x^2+9x+7)<=1+(-8)/43=35/43`
Vậy `Q_(max)=35/43`