How do you combine #(x-1)*[3/(x^2-1)+x/(2x-2)]#?

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Answer 1 · 2017-08-22T22:29:25

#=(x^2+x+6)/(2(x+1))#

#(x-1)*[3/(x^2-1)+x/(2x-2)]#

Factorise the denominators first

#= (x-1)*[3/((x+1)(x-1))+x/(2(x-1))]#

Add the fractions using a common denominator.

#=(x-1)*[((2xx3 + x(x+1))/(2(x+1)(x-1))]#

#=(x-1)*[(6 + x^2+x)/(2(x+1)(x-1))]#

Distribute #(x-1)# into the bracket:

#=[(cancel((x-1))(6 + x^2+x))/(2(x+1)cancel((x-1))]#

#=(x^2+x+6)/(2(x+1))#