a) `f(x)=(x+1)(8-5x)`
`f(x)=-5x^2+3x+8`
`f(x)=-5(x^2-3/5x+9/100)+169/20`
`f(x)=-5(x-3/10)^2+169/20\le 169/20`
Dấu `=` xảy ra `⇔x-3/10=0⇔x=3/10`
Vậy $Max_{f(x)}=\dfrac{169}{20}⇔x=\dfrac{3}{10}$
b) `f(x)=3x^2(8-x^2)`
`f(x)=-3x^4+24x^2`
`f(x)=-3(x^4-8x^2+16)+48`
`f(x)=-3(x^2-4)^2+48\le 48`
Dấu `=` xảy ra `⇔x^2-4=0⇔x=±2`
Vậy $Max_{f(x)}=48⇔x=±2$