Đáp án :
`a)ĐKXĐ : x \ne -3; x \ne 1`
`A=(x^2+16)/(x+3)`
`b)A_(min)=4` khi `x=2`
Giải thích các bước giải :
`a)ĐKXĐ : x \ne -3; x \ne 1`
`A=(x^3+26x-19)/((x-1)(x+3))+(2x)/(1-x)+(x-3)/(x+3)`
`<=>A=(x^3+26x-19)/((x-1)(x+3))-(2x)/(x-1)+(x-3)/(x+3)`
`<=>A=(x^3+26x-19)/((x-1)(x+3))-(2x(x+3))/((x-1)(x+3))+((x-3)(x-1))/((x+3)(x-1))`
`<=>A=(x^3+26x-19)/((x-1)(x+3))-(2x^2+6x)/((x-1)(x+3))+(x^2-4x+3)/((x-1)(x+3))`
`<=>A=(x^3+26x-19-2x^2-6x+x^2-4x+3)/((x-1)(x+3))`
`<=>A=(x^3-x^2+16x-16)/((x-1)(x+3))`
`<=>A=(x^2(x-1)+16(x-1))/((x-1)(x+3))`
`<=>A=((x-1)(x^2+16))/((x-1)(x+3))`
`<=>A=(x^2+16)/(x+3)`
`b)A=(x^2+16)/(x+3)`
`<=>A-4=(x^2+16)/(x+3)-4`
`<=>A-4=(x^2+16)/(x+3)-(4x+12)/(x+3)`
`<=>A-4=(x^2-4x+16-12)/(x+3)`
`<=>A-4=(x^2-4x+4)/(x+3)`
`<=>A-4=(x-2)^2/(x+3) ≥ 0`
`=>A-4 ≥ 0`
`=>A ≥ 4`
`=>A_(min)=4`
Xảy ra dấu `=` khi :
`(x-2)^2/(x+3)=0`
`<=>x-2=0`
`<=>x=2` (Thỏa mãn : `x \ne 3; x \ne 1; x> -3`)
Vậy : `A_(min)=4` khi `x=2`
