Đáp án:
Giải thích các bước giải:
a,
`A=4x-x^2-11`
`=-x^2+4x-11`
`=-x^2+4x-4-7`
`=-(x^2-4x+4)-7`
`=-(x-2)^2-7<=-7`
Dấu `=` xảy ra `<=>x-2=0<=>x=2`
Vậy `A_(max)=7<=>x=2`
b,
`A=4x^2-6xy+9y^2-16x+12y+2012`
`=(9y^2-6xy+12y)+4x^2-16x+2012`
`=[9y^2-2·3y·(x-2)+(x-2)^2]-(x-2)^2+4x^2-16x+2012`
`=[3y-(x-2)]^2-x^2+4x-4+4x^2-16x+2012`
`=(3y-x+2)^2+3x^2-12x+2008`
`=(3y-x+2)^2+3x^2-12x+12+1996`
`=(3y-x+2)^2+3(x-2)^2+1996>=1996`
Dấu `=` xảy ra `<=>`$\left \{ {{x-2=0} \atop {3y=x-2}} \right.$
`<=>`$\left \{ {{x=2} \atop {3y=0}} \right.$
`<=>`$\left \{ {{x=2} \atop {y=0}} \right.$
Vậy `A_(min)=1996<=>x=2; y=0`
