Question

Answer 1

`A=((x^{2}-2006x+2009)/(x^{2}-1)-(x)/(x-1)+(2008)/(x+1)):(x^{2}+3x)/(x^{2}-x)\   (ĐK:x\ne {+-1;0;-3}`

`=((x^{2}-2006x+2009)/((x-1)(x+1))-(x)/(x-1)+(2008)/(x+1)):(x(x+3))/(x(x-1))`

`=(x^{2}-2006x+2009-(x+1)+2008(x-1))/((x-1)(x+1)).(x(x-1))/(x(x+3))`

`=(x^{2}-2006x+2009-x-1+2008x-2008)/((x-1)(x+1)).(x(x-1))/(x(x+3))`

`=(x^{2}+x)/((x-1)(x+1)).(x(x-1))/(x(x+3))`

`=(x(x+1))/((x-1)(x+1)).(x(x-1))/(x(x+3))`

`=(x)/(x+3)`

Answer 2

`A=((x^{2}-2006x+2009)/((x-1)(x+1))-(x)/(x-1)+(2008)/(x+1)):(x(x+3))/(x(x-1))\   (ĐK:x\ne{+-1;0;-3})`

`=(x^{2}-2006x+2009-x(x+1)+2008(x-1))/((x+1)(x-1)).(x(x-1))/(x(x+3))`

`=(x^{2}-2006x+2009-x^{2}-x+2008x-2008)/((x+1)(x-1)).(x(x-1))/(x(x+3))`

`=(x+1)/((x+1)(x-1)).(x(x-1))/(x(x+3))`

`=(1)/(x+3)`