`A=1/(x-1)-1/(x+1)+(4x^2+2)/(x^2-1)`
`đk:x ne +-1`
`a)A=(x+1-x+1+4x+2)/(x^2-1)`
`=(4x+4)/(x^2-1)`
`b)A=4/2015`
`<=>(4x+4)/(x^2-1)=4/2015`
`<=>(x+1)/(x^2-1)=1/2015`
`<=>(x+1)/((x-1)(x+1))=1/2015`
`<=>1/(x-1)=1/2015`
`<=>x-1=2015`
`<=>x=2016`
Vậy `x=2016` thì `A=4/2015`