`a) A=4x-x^2`
`A=-(x^2-4x)`
`A=-(x^2-4x+4)+4`
`A=-(x-2)^2+4`
Vì `-(x-2)^2` $\leq$ `0` `∀x`
`=> A=-(x-2)^2+4` $\leq$ `4`
Dấu = xảy ra `<=> -(x-2)^2=0`
`=> x-2=0`
`=> x=2`
Vậy `A_(max)=4 <=> x=2`
`b) B=x^2+5x+12`
`B=(x^2-5x+25/4)+23/4`
`B=(x-5/2)^2+23/4` $\leq$ `23/4`
Dấu = xảy ra `<=> (x-5/2)^2=0`
`=> x=5/2`
Vậy `B_(min)=23/4 <=> x=5/2`