Đáp án:
`F_(max)=40/3` khi `x=-7/3`
Giải thích các bước giải:
`F=(x-1)^2-(2x+3)^2+5`
`=x^2-2x+1-[(2x)^2+2.2x.3+3^2]+5`
`=x^2-2x+1-(4x^2+12x+9)+5`
`=x^2-2x+1-4x^2-12x-9+5`
`=(x^2-4x^2)+(-2x-12x)+(1-9+5)`
`=-3x^2-14x-3`
`=-3(x^2+14/3x+1)`
`=-3(x^2+14/3x+49/9-40/9)`
`=-3(x^2+14/3+49/9)+40/3`
`=-3[x^2+2.x. 7/3+(7/3)^2]+40/3`
`=-3(x+7/3)^2+40/3`
Ta có:`(x+7/3)^2≥0∀x`
`⇒3(x+7/3)^2≥0∀x`
`⇒-3(x+7/3)^2≤0∀x`
`⇒-3(x+7/3)^2+40/3≤40/3∀x`
Vậy `F_(max)=40/3` khi `x+7/3=0⇔x=-7/3`
