`1.`
`A=2x²-x-1`
`=2(x²-1/2x-1/2)`
`=2(x²-1/2x+1/16-9/16)`
`=2(x²-1/2x+1/16)-9/8`
`=2[x²-2.x. 1/4+(1/4)^2]-9/8`
`=2(x-1/4)^2-9/8`
Ta có:`(x-1/4)^2≥0∀x`
`⇒2(x-1/4)^2≥0∀x`
`⇒2(x-1/4)^2-9/8≥-9/8∀x`
Vậy `A_(min)=-9/8` khi `x-1/4=0⇔x=1/4`
`B=3x²+6x-3`
`=3(x²+2x-1)`
`=3(x²+2x+1-2)`
`=3(x²+2x+1)-6`
`=3(x+1)²-6`
Ta có:`(x+1)²≥0∀x`
`⇒3(x+1)²≥0∀x`
`⇒3(x+1)²-6≥-6∀x`
Vậy `B_(min)=-6` khi `x+1=0⇔x=-1`
`2.`
`A=2x-x²-4`
`=-(x²-2x+4)`
`=-(x²-2x+1+3)`
`=-(x²-2x+1)-3`
`=-(x-1)²-3`
Ta có:`(x-1)²≥0∀x`
`⇒-(x-1)²≤0∀x`
`⇒-(x-1)²-3≤-3∀x`
Vậy `A_(max)=-3` khi `x-1=0⇔x=1`
`B=3-2x²+4x`
`=-2(x²-2x-3/2)`
`=-2(x²-2x+1-5/2)`
`=-2(x²-2x+1)+5`
`=-2(x-1)²+5`
Ta có:`(x-1)²≥0∀x`
`⇒2(x-1)²≥0∀x`
`⇒-2(x-1)²≤0∀x`
`⇒-2(x-1)²+5≤5∀x`
Vậy `B_(max)=5` khi `x-1=0⇔x=1`
