`A= x^2 + 5y^2 - 2xy + 4y + 3`
`A= x^2 - 2xy+ y^2 +4y^ +4y + 1+2`
`A=(x^2 - 2xy+ y^2 ) +(4y^ +4y + 1)+2`
`A=(x-y )^2 +(2y+1)^2+2`
Có: `(x-y )^2\ge0, (2y+1)^2\ge0`
`⇒A\ge0+0+2=2.`
Dấu ''='' xảy ra khi `⇔`$\begin{cases}(x-y )^2=0\\(2y+1)^2=0\\\end{cases}$
`⇔x=y=-1/2.`
Vậy $MinA=2$`⇔x=y=-1/2.`
`b) B = ( x^2 - 2x ) ( x^2 - 2x + 2 )`
`B = ( x^2 - 2x ) [( x^2 - 2x) + 2 )]`
`B=(x^2-2x)^2+2(x^2-2x)+1-1`
`B=(x^2-2x+1)^2-1`
`B=((x-1)^2)^2-1`
`B=(x-1)^4-1`
Có: `(x-1)^4\ge0`
`⇒A\ge0-1=-1.`
Dấu ''='' xảy ra khi `x=1.`
Vậy $MinC=-1$`⇔x=1.`
`c)C = x^2 - 4xy + 5y^2 + 10x - 22y + 28`
`C = x^2 - 4xy + 10x + 4y^2 - 20y + 25 + y^2 - 2y + 1 + 2`
`C = x^2 - (4xy - 10x) + (4y^2 - 20y + 25) +( y^2 - 2y + 1 )+ 2`
`C = x^2 - 2x(2y-5) +(2y-5)^2 + (y-1)^2 + 2`
`C = (x-2y+5)^2 + (y-1)^2 + 2`
Có: ` (x-2y+5)^2\ge0; (y-1)^2\ge0 `
`⇒C\ge0+0+2=2.`
Dấu ''='' xảy ra khi `⇔`$\begin{cases}(x-2y+5)^2=0\\(y-1)^2=0\\\end{cases}$`
`⇔`$\begin{cases}x=-3\\y=1\\\end{cases}$`
Vậy $MinC=2$`⇔$\begin{cases}x=-3\\y=1\\\end{cases}$.`
