Đáp án:
`A_(min)=-7 <=> x=2`
`B_(max)=9/5 <=> x=-2/5`
Giải thích các bước giải:
`a)`
`A=2x^2-8x+1`
`=2x^2-8x+8-7`
`=2.(x^2-4x+4)-7`
`=2.(x-2)^2-7`
Ta có : `(x-2)^2>=0 to 2.(x-2)^2>=0`
`to 2.(x-2)^2-7>=-7`
Dấu "=" xảy ra `<=> (x-2)^2=0`
`=> x=2`
Vậy `A_(min)=-7 <=> x=2`
`b)`
`B=-5x^2-4x+1`
`= -5 . (x^2+4/5x-1/5) `
`= -5 . [x^2+2.x . 2/5+(2/5)^2-(2/5)^2-1/5]`
`=-5.[(x+2/5)^2-9/25]`
`=-5.(x+2/5)^2+9/5`
Ta có : `(x+2/5)^2>=0 to -5.(x+2/5)^2<=0`
`to -5.(x+2/5)^2+9/5<=9/5`
Dấu "=" xảy ra `<=> (x+2/5)^2=0`
`=> x=-2/5`
Vậy `B_(max)=9/5 <=> x=-2/5`
