Đáp án:
`A_(min)=2020 <=> x=2, y=2, z=1`
Giải thích các bước giải:
`A=x^2+y^2+z^2-yz-4x-3y+2027`
`4A=4x^2+4y^2+4z^2-4yz-16x-12y+8108`
`=(4x^2-16x+16)+(3y^2-12y+12)+(y^2-4yz+4z^2)+8080`
`=4(x^2-4x+4)+3(y^2-4y+4)+(y-2z)^2+8080`
`=4(x-2)^2+3(y-2)^2+(y-2z)^2+8080`
Ta có: $\left\{\begin{matrix}4(x-2)^2≥0& \\3(y-2)^2≥0&\\(y-2z)^2 ≥0& \end{matrix}\right.$
`=> 4(x-2)^2+3(y-2)^2+(y-2z)^2>=0`
`=> 4(x-2)^2+3(y-2)^2+(y-2z)^2+8080>=8080`
`=> A>=2020`
Dấu "=" xảy ra `<=>` $\left\{\begin{matrix}4(x-2)^2=0& \\3(y-2)^2=0&\\(y-2z)^2 =0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x=2& \\y=2 \\y=2z& \end{matrix}\right.$`=>` $\left\{\begin{matrix}x=2& \\y=2 \\z=1& \end{matrix}\right.$
Vậy `A_(min)=2020 <=> x=2, y=2, z=1`
