Đáp án:
a)ĐK:\(\begin{cases}x^2-1 \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ne 0\\x \ne 1\\x \ne -1\\\end{cases}\)
`A=((x^3+1)/(x^2-1)-(x^2-1)/(x-1)):(x+x/(x-1)`
`=(((x+1)(x^2-x+1))/((x-1)(x+1))-(x^2-1)/(x-1)):((x^2-x+x)/(x-1))`
`=((x^2-x+1)/(x-1)-(x^2-1)/(x-1))*(x-1)/x^2`
`=((x^2-x+1-x^2+1)/(x-1))*(x-1)/x^2`
`=(2-x)/(x-1)*(x-1)/x^2`
`=(2-x)/x^2`
`b)A=3`
`<=>(2-x)/x^2=3`
`<=>2-x=3x^2`
`<=>3x^2+x-2=0`
`<=>3x^2+3x-2x-2=0`
`<=>3x(x+1)-2(x+1)=0`
`<=>(x+1)(3x-2)=0`
Vì `x ne -1<=>x+1 ne 0`
`<=>3x-2=0`
`<=>x=2/3`
`c)A in ZZ`
`=>2-x vdots x^2`
`=>(2-x)(2+x) vdots x^2`(nhân 2+x)
`=>4-x^2 vdots x^2`
`=>4 vdots x^2`
`=>x^2 in Ư(4)={+-1,+-2,+-4}`
`=>x^2 in {1,2,4}(do \ x^2>=0)`
`=>x in {+-1,+-2}(do \ x in ZZ)`
`=>x in {2,-2}(do \ đkxđ:x ne +-1)`
Thử lại ta thấy `x=2` và `x=-2` thỏa mãn.
Vậy `x in {2,-2}` thì `A in ZZ.`
