`B= (( x+1)/ (2x -2) +3/(x^2-1) - (x+3)/(2x+2)) . (4x^2 - 4)/5`
$a)ĐKXĐ:\begin{cases}2x-2\ne0\\x^2\ne0\\2x+2\ne0\\\end{cases}$
`<=>`$\begin{cases}x\ne1\\x\ne-1\end{cases}$
`b)B= (( x+1)/ (2(x -1)) +3/((x-1)(x+1)) - (x+3)/(2(x+1))) . (4x^2 - 4)/5`
`B= (( x+1)^2/ (2(x -1)(x+1)) +3.2/(2(x-1)(x+1)) - ((x+3)(x-1))/(2(x+1)(x-1))) . (4x^2 - 4)/5`
`B= (( x+1)^2 +3.2 - (x+3)(x-1))/(2(x+1)(x-1)) . (4x^2 - 4)/5`
`B= (x^2+2x+1 +6 - (x^2+2x-3))/(2(x+1)(x-1)) . (4x^2 - 4)/5`
`B= (x^2+2x+1 +6 - x^2-2x+3)/(2(x+1)(x-1)) . (4x^2 - 4)/5`
`B= (10)/(2(x+1)(x-1)) . (4x^2 - 4)/5`
`B= (10)/(2(x^2-1)) . (4(x^2 - 1))/5`
`B= 4`
