a) `ĐK: x`$\neq±1$
`A=[(x+1)/(1-x)-(1-x)/(1+x)-(4x^2)/(x^2-1)]:(4x^2-4)/(x^2-2x+1)`
`=[(x+1)^2-(1-x)^2+4x^2]/(1-x)(1+x):[4(x-1)(x+1)]/(x-1)^2`
`=[4x(x+1)/(1-x)(1+x).(x-1)/4(x+1)`
`=-x/(x+1)`
b) `A=-3⇔-x/x+1=-3⇔-3(x+1)=-x⇔-3x-3=-x⇔-2x=3⇔x=-3/2`
Vậy `x=-3/2` thì `A=-3`
c) `A=-x/(x+1)=[-(x+1)+1]/(x+1)=-1+1/(x+1)∈Z`
`⇔1/(x+1)∈Z⇒x+1∈Ư(1)=\{±1\}`
`⇒x∈\{0;-2\}` (tmđk)
Vậy `x∈\{0;-2\}`