`A=8x-8x²+3`
`= -8(x²-x-3/8)`
`=-8(x²-x+1/4-5/8)`
`=-8(x²-x+1/4)+5`
`=-8[x²-2.x. 1/2+(1/2)^2]+5`
`=-8(x-1/2)^2+5`
Ta có:`(x-1/2)^2≥0∀x`
`⇒8(x-1/2)^2≥0∀x`
`⇒-8(x-1/2)^2≤0∀x`
`⇒-8(x-1/2)^2+5≤5∀x`
Vậy `A_(max)=5` khi `x-1/2=0⇔x=1/2`
$\\$
`C=12x-4x²+3`
`=-4(x²-3x-3/4)`
`=-4(x²-3x+9/4-3)`
`=-4(x²-3x+9/4)+12`
`=-4[x²-2.x. 3/2+(3/2)^2]+12`
`=-4(x-3/2)^2+12`
Ta có:`(x-3/2)^2≥0∀x`
`⇒4(x-3/2)^2≥0∀x`
`⇒-4(x-3/2)^2≤0∀x`
`⇒-4(x-3/2)^2+12≤12∀x`
Vậy `C_(max)=12` khi `x-3/2=0⇔x=3/2`
