Bài 6:
`1, M=x^2+y^2-2x+4y+2025`
`M=(x^2-2x+1)+(y^2+4y+4)+2020`
`M=(x-1)^2+(y+2)^2+2020>=2020`
Dấu = xảy ra khi `x-1=0; y+2=0`
`<=> x=1; y=-2`
Vậy `M_min=2020` khi `x=1; y=-2`
`2,` Ta có:
`4x^2+4x+3=(4x^2+4x+1)+2`
`=(2x+1)^2+2>=2`
`=> 4040/(4x^2+4x+3)<=4040/2=2020`
`=> A<=2020`
Dấu = xảy ra khi `2x+1=0`
`<=> x=-1/2`
Vậy `A_max=2020` khi `x=-1/2`