Toán Tìm $Max_{Q}=\dfrac{3x^{2}+9x+17}{3x^{2}+9x+7}$ 29/11/2021 By Cora Tìm $Max_{Q}=\dfrac{3x^{2}+9x+17}{3x^{2}+9x+7}$
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` Trả lời
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`
Đáp án:
Giải thích các bước giải:
$Q<=>1-$ $\frac{10}{3x²+9x+7}$
Vậy $Q_{max}$$=1$