`a)x²+x+1`
`=x²+x+1/4+3/4`
`=(x²+x+1/4)+3/4`
`=[x²+2.x. 1/2+(1/2)^2]+3/4`
`=(x+1/2)^2+3/4`
Ta có:`(x+1/2)^2≥0∀x`
`⇒(x+1/2)^2+3/4≥3/4∀x`
Vậy `x²+x+1_(min)=3/4` khi `x+1/2=0⇔x=-1/2`
`b)2+x-x²`
`=-(x²-x-2)`
`=-(x²-x+1/4-9/4)`
`=-(x²-x+1/4)+9/4`
`=-[x²-2.x. 1/2+(1/2)^2]+9/4`
`=-(x-1/2)^2+9/4`
Ta có:`(x-1/2)^2≥0∀x`
`⇒-(x-1/2)^2≤0∀x`
`⇒-(x-1/2)^2+9/4≤9/4∀x`
Vậy `2+x-x²_(max)=9/4` khi `x-1/2=0⇔x=1/2`
`c)x²-4x+1`
`=x²-4x+4-3`
`=(x²-4x+4)-3`
`=(x²-2.x.2+2²)-3`
`=(x-2)²-3`
Ta có:`(x-2)²≥0∀x`
`⇒(x-2)²-3≥-3∀x`
Vậy `x²-4x+1_(min)=-3` khi `x-2=0⇔x=2`
`d)4x²+4x+11`
`=4x²+4x+1+10`
`=(4x²+4x+1)+10`
`=[(2x)²+2.2x.1+1²]+10`
`=(2x+1)²+10`
Ta có:`(2x+1)²≥0∀x`
`⇒(2x+1)²+10≥10∀x`
Vậy `4x²+4x+11_(min)=10` khi `2x+1=0⇔x=-1/2`
`e)3x²-6x+1`
`=3(x²-2x+1/3)`
`=3(x²-2x+1-2/3)`
`=3(x²-2x+1)-2`
`=3(x-1)²-2`
Ta có:`(x-1)²≥0∀x`
`⇒3(x-1)²≥0∀x`
`⇒3(x-1)²-2≥-2∀x`
Vậy `3x²-5x+1_(min)=-2` khi `x-1=0⇔x=1`
`f)x²-2x+y²-4y+6`
`=x²-2x+y²-4y+1+4+1`
`=(x²-2x+1)+(y²-4y+4)+1`
`=(x²-2.x.1+1²)+(y²-2.y.2+2²)+1`
`=(x-1)²+(y-2)²+1`
Dấu `'='` xảy ra khi$\begin{cases} x-1=0\\y-2=0 \end{cases}$`⇔`$\begin{cases} x=1\\y=2 \end{cases}$
Vậy `x²-2x+y²-4y+6_(min)=1` khi `x=1` và `y=2`
