Đáp án:
`MIN_B=-1/4 \harr x=3/2`
Giải thích các bước giải:
`A=(2x+1)/((x-4)(x-3))-(x+3)/(x-4) +(2x+1)/(x-3)`
`ĐKXĐ:x ne 3,4`
`->A=(2x+1-(x+3)(x-3)+(2x+1)(x-4))/((x-3)(x-4))`
`->A=(2x+1-x^2+9+2x^2-7+4)/((x-3)(x-4))`
`->A=(x^2-5x+6)/((x-3)(x-4))`
`->B=A.(x^2-5x+4)=((x^2-5x+6)(x^2-5x+4))/((x-3)(x-4))`
`->B=((x-2)(x-3)(x-1)(x-4))/((x-3)(x-4))`
`->B=(x-1)(x-2)`
`->B=x^2-3x+2`
`->B=x^2-2.x.3/2+9/4-1/4`
`->B=(x-3/2)^2-1/4>=-1/4`
Dấu = xảy ra khi `x-3/2=0->x=3/2`
Vậy `MIN_B=-1/4 \harr x=3/2`
`cancel{nocopy//2072007}`