How do you add #(a-1)/(a+1 )+ (a+1)/(a-1)#?

1 Answer
Alan P.
Oct 2, 2015

Convert the denominators to a common value and simplify to get
#color(white)("XXX")(2(a^2+1))/(a^2-1)#

Explanation:

#color(red)((a-1)/(a+1)) + color(blue)((a+1)/(a-1))#

#color(white)("XXX")=color(red)((a-1)/(a+1)) * (a-1)/(a-1) + (color(blue)(a+1)/(a-1)) * (a+1)/(a+1)#

#color(white)("XXX")=(color(green)((a-1)^2)+color(brown)((a+1)^2))/((a-1)(a+1)#

#color(white)("XXX")=(color(green)(a^2-2a+1)+color(brown)(a^2+2a+1))/((a-1)(a+1))#

#color(white)("XXX")=(2(a^2+1))/(a^2-1)#

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