Question

Tính GTLN của A=15/(10+(x+1)(x-4)) +2017

Answer 1

\(A=\dfrac{15}{10+\left(x+1\right)\left(x-4\right)}+2017\)

\(=\dfrac{15}{10+x^2-3x-4}+2017\)

\(=\dfrac{15}{x^2-3x+6}+2017\)

Có \(x^2-3x+6=x^2-2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{15}{4}=\left(x-\dfrac{3}{2}\right)+\dfrac{15}{4}\ge\dfrac{15}{4}\)

\(\Leftrightarrow A\le2021\)

Dấu = \(\Leftrightarrow x=\dfrac{3}{2}\)

Answer 2

`A=15/(10+(x+1)(x-4))+2017`

`=15/(x^2-3x-4+10)+2017`

`=15/(x^2-3x+6)+2017`

Vì `x^2-3x+6`

`=(x-3/2)^2+15/4>=15/4`

`=>15/(x^2-3x+6)<=15:15/4=4`

`=>A<=2017+4=2021`

Dấu “=” `<=>x=3/2`