Đáp án + Giải thích các bước giải:
`a//` `ĐKXĐ:x\ne±1`
`C=((x-2)/(2x-2)-(3)/(2x+2)+(x)/(x^{2}-1)):(1-x)/(2x)`
`=>C=((x-2)/(2(x-1))-(3)/(2(x+1))+(x)/((x-1)(x+1))).(2x)/(1-x)`
`=>C=((x-2)(x+1)-3(x-1)+2x)/(2(x-1)(x+1)).(2x)/(1-x)`
`=>C=(x^{2}-2x+x-2-3x+3+2x)/(2(x-1)(x+1)).(2x)/(1-x)`
`=>C=(x^{2}-2x+1)/(2(x-1)(x+1)).(2x)/(1-x)`
`=>C=((x-1)^{2})/(2(x-1)(x+1)).(2x)/(1-x)`
`=>C=(x-1)/(2(x+1)).(2x)/(1-x)`
`=>C=(2x(x-1))/(2(x+1)(1-x))`
`=>C=-(2x(x-1))/(2(x+1)(x-1))`
`=>C=-(x)/(x+1)`
`b//`
`C=-(x)/(x+1)∈ZZ`
`=>x\vdots x+1`
`=>(x+1)-1\vdots x+1`
Vì `(x+1)\vdots x+1`
`=>1\vdots x+1`
`=>x+1∈Ư(1)={±1}`
`=>x∈{0;-2}\ \ \ (TM)`
`c//`
`C<0`
`<=>-(x)/(x+1)<0`
`<=>(x)/(x+1)>0`
`+)TH1:`
$\left\{\begin{matrix}x>0& \\x+1>0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x>0& \\x> -1& \end{matrix}\right.$
`=>x>0`
`+)TH2:`
$\left\{\begin{matrix}x<0& \\x+1<0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x<0& \\x< -1& \end{matrix}\right.$
`=>x< -1`
Vậy với `x>0\ \;\ \x< -1` và `x\ne1` thì `C<0`
