Lời giải.
`A = {2x^2-2x+3}/{x^2-x+2}`
`A={2(x^2-x+2)-1}/{x^2-x+2}`
`A={2(x^2-x+2)}/{x^2-x+2}-1/{x^2-x+2}`
`A=2-1/{x^2-x+2}`
Ta có: `x^2-x+2=x^2 - 2. x . 1/2 + 1/4 + 7/4 = (x-1/2)^2+7/4≥7/4`
`=>1/{x^2-x+2}≤1/{frac{3}{4}}=4/7`
`=>-1/{x^2-x+2}≥-4/7`
`=>A=2+(-1/{x^2-x+2})≥-4/7 +2=-4/7+14/7=10/7.`
Dấu "=" xảy ra `<=>x-1/2=0<=>x=1/2.`
Vậy `minA=10/7` khi `x=1/2.`
