`**` SỬA ĐỀ: `x^2-mx+m-4=0`

`m^2-4(m-4)>=0`

`m^2-4m+16>=0`

`m inR`

`{(x_1+x_2=m),(x_1*x_2=m-4):}`

Ta có:

`A=3(x_1^2+x_2^2)-2x_1x_2`

`A=3[(x_1^2+2x_1x_2+x_2^2)-2x_1x_2]-2x_1x_2`

`A=3[(x_1+x_2)^2-2x_1x_2]-2x_1x_2`

`A=3[m^2-2(m-4)]-2(m-4)`

`A=3(m^2-2m+8)-2m+8`

`A=3m^2-6m+24-2m+8`

`A=3m^2-8m+32`

`A=(3m^2-8m+16/3)+32-16/3`

`A=(sqrt3m-(4sqrt3)/3)^2+80/3`

Vì `(sqrt3m-(4sqrt3)/3)^2>=0 AAx`

nên `(sqrt3m-(4sqrt3)/3)^2+80/3>=80/3 AAx`

Hay `A>=80/3 AAx` 

Dấu `"="` xảy ra khi: `(sqrt3m-(4sqrt3)/3)^2=0`

`sqrt3m-(4sqrt3)/3=0`

`sqrt3m=(4sqrt3)/3`

`m=4/3`

Vậy `M i n _A=80/3` khi `m=4/3`