`a)C=((2x^2+1)/(x^3-1)-1/(x-1)):(1-(x^2-2)/(x^2+x+1))`
`ĐK:x ne 1`
`C=((2x^2+1-x^2-x-1)/(x^3-1)):((x^2+x+1-x^2+2)/(x^2+x+1))`
`C=((x^2-x)/(x^3-1)):((x+3)/(x^2+x+1))`
`C=x/(x^2+x+1)*(x^2+x+1)/(x+3)`
`C=x/(x+3)`
`b)|1-x|+2=3(x+1)`
`<=>|1-x|+2=3x+3`
`<=>|1-x|=3x+1(x>=-1/3)`
`**1-x=3x+1`
`<=>4x=0<=>x=0(tmđk)`
`**x-1=3x+1`
`<=>2x=-2`
`<=>x=-1(l)`
Thay `x=0` vào C
`=>C=0`
`c)C in ZZ`
`=>x vdots x+3`
`=>x+3-3 vdots x+3`
`=>3 vdots x+3`
`=>x+3 in Ư(3)={+-1,+-3}`
`=>x in {-2,-4,0,-6}`
`d)|C|>C`
Mà `|C|>=0`
`=>C<0`
`<=>x/(x+3)<0`
Để 1 p/s `<=0` thì tử và mẫu trái dấu mà `x<x+3`
`=>` \(\begin{cases}x<0\\x+3>0\\\end{cases}\)
`<=>` \(\begin{cases}x>-3\\x<0\\\end{cases}\)
`<=>-3<x<0`
