Question 131902
{{{6/(x^2-4x-5) + 1/(x+1)}}}


The first step is to factor the denominator in the first fraction.


{{{-5*1=-5}}} and {{{-5+1=-4}}}, so:


{{{x^2-4x-5=(x-5)(x+1)}}}


And Ah ha! The two denominators have a common factor of {{{x+1}}}, so:


{{{6/(x^2-4x-5) + 1/(x+1)=6/((x-5)(x+1)) + (1/(x+1))((x-5)/(x-5))=(6+(x-5))/((x-5)(x+1))=(x+1)/((x-5)(x+1))=1/(x-5)}}}


That would be D.  Final Answer, Regis