Đáp án:
`1)x^2+5y^2+9z^2-4xy-6yz+12`
`=x^2-4xy+4y^2+y^2-6yz+9z^2+12`
`=(x-2y)^2+(y-3z)^2+12>=12AAx,y,z in RR`
Dấu "=" xảy ra khi \(\begin{cases}x-2y=0\\y-3z=0\\\end{cases}\)
`<=>` \(\begin{cases}x=2y\\y=3z\\\end{cases}\)
`2)x^2+6y^2+z^2-4xy-7y+2yz+2x+2023`
`=x^2-4xy+4y^2+2y^2+2yz+z^2+2x-7y+2023`
`=(x-2y)^2+2(x-2y)+1+y^2+2yz+z^2+y^2-3y+2022`
`=(x-2y+1)^2+(y+z)^2+y^2-3y+9/4+2022-2-1/4`
`=(x-2y+1)^2+(y+z)^2+(y-3/2)^2+2019 3/4>=2019 3/4`
Dấu "=" xảy ra khi \(\begin{cases}x-2y+1=0\\y+z=0\\y-\dfrac32=0\\\end{cases}\)
`<=>` \(\begin{cases}y=\dfrac32\\z=-y=-\dfrac32\\x=2y-1=2\\\end{cases}\)
`3)(x-a)^2+(x-b)^2+(x-c)^2`
Vì \(\begin{cases}(x-a)^2\ge0\\(x-b)^2\ge0\\(x-c)^2\ge0\\\end{cases}\)
`=>(x-a)^2+(x-b)^2+(x-c)^2\ge0`
Dấu"=" xảy ra khi \(\begin{cases}x=a\\x=b\\x=c\\\end{cases}\)
`<=>a=b=c=x`
