Lời giải:
`a)`
`M=x^(2)-3x+10`
`=x^(2)-3x+9/4+31/4`
`=(x^(2)-3x+9/4)+31/4`
`=[x^(2)-2.x.(3)/2+(3/2)^2]+31/4`
`=(x-3/2)^(2)+31/4`
Ta có `(x-3/2)^(2)+31/4≥31/4`
Dấu "`=`" xảy ra khi :
`(x-3/2)^2=0`
`⇔x-3/2=0`
`⇔x=3/2`
Vậy `text{MinM}=31/4` khi `x=3/2`
`b)`
`A=25x^(2)-20x+7`
`=25x^(2)-20x+4+3`
`=[(5x)^(2)-2.5x.2+2^2]+3`
`=(5x-2)^(2)+3`
Ta có `(5x-2)^(2)+3≥0`
Dấu "`=`" xảy ra khi :
`(5x-2)^2=0`
`⇔5x-2=0`
`⇔x=2/5`
Vậy `text{MinA=3}` khi `x=2/5`
`c)`
`C=x^(2)-8x+20`
`=x^(2)-8x+16+4`
`=(x^(2)-2.x.4+4^2)+4`
`=(x-4)^(2)+4`
Ta có `(x-4)^(2)+4≥0`
Dấu `=` xảy ra khi :
`(x-4)^2=0`
`⇔x-4=0`
`⇔x=4`
Vậy `text{MinC =4}` khi `x=4`
`d)`
`D=4x^(2)-12x+11`
`=4x^(2)-12x+9+2`
`=[(2x)^(2)-2.2x.3+3^2]+2`
`=(2x-3)^(2)+2`
Ta có : `(2x-3)^2+2≥2`
Dấu "`=`" xảy ra khi :
`(2x-3)^2=0`
`⇔2x-3=0`
`⇔x=3/2`
Vậy `text{MinA=2}` khi `x=3/2`
