Đáp án:
Giải thích các bước giải:
`A=2x^2+x+1`
`A=2(x^2+1/2 x+1/2)`
`A=2(x^2+2 . 1/4 x+1/16+7/16)`
`A=2[(x+1/4)^2+7/16]`
`A=2(x+1/4)^2+7/8`
Ta có: `2(x+1/4)^2 \ge 0 \forall x`
`⇒ 2(x+1/4)^2+7/8 \ge 7/8 \forall x`
Vậy `A_{min}=7/8` khi `x+1/4=0⇔x=-1/4`
`B=3x^2+5x+2`
`B=3(x^2+5/3 x+2/3)`
`B=3(x^2+2 . 5/6 x+25/36-1/36)`
`B=3[(x+5/6)^2-1/36]`
`B=3(x+5/6)^2-1/12`
Ta có: `3(x+5/6)^2 \ge 0 \forall x`
`⇒ 3(x+5/6)^2-1/12 \ge -1/12 \forall x`
Vậy `B_{min}=-1/12` khi `x+5/6=0⇔x=-5/6`
