`a)A=x²+4x+7`
`=(x²+4x+4)+3`
`=(x²+2.x.2+2²)+3`
`=(x+2)²+3`
Ta có:`(x+2)²>=0AAx`
`⇒(x+2)²+3>=3AAx`
Vậy `A_(min)=3` khi `x+2=0⇔x=-2`
`b)B=x²-x+1`
`=(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>=0AAx`
`⇒(x-1/2)^2+3/4>=3/4AAx`
Vậy `B_(min)=3/4` khi `x-1/2=0<=>x=1/2`
`c)C=4x-x²+3`
`=-(x²-4x-3)`
`=-(x²-4x+4-7)`
`=-(x²-4x+4)+7`
`=-(x²-2.x.2+2²)+7`
`=-(x-2)²+7`
Ta có:`(x-2)²>=0AAx`
`⇒-(x-2)²<=0AAx`
`⇒-(x-2)²+7<=7AAx`
Vậy `C_(max)=7` khi `x-2=0⇔x-2`
`d)D=2x-2x²-5`
`=-2(x²-x+5/2)`
`=-2(x²-x+1/4+9/4)`
`=-2(x²-x+1/4)-9/2`
`=-2[x²-2.x. 1/2+(1/2)^2]-9/2`
`=-2(x-1/2)^2-9/2`
Ta có:`(x-1/2)^2>=0AAx`
`⇒2(x-1/2)^2>=0AAx`
`⇒-2(x-1/2)^2<=0AAx`
`⇒-2(x-1/2)^2-9/2<=-9/2AAx`
Vậy `D_(max)=-9/2` khi `x-1/2=0⇔x=1/2`
