How do you evaluate #\frac { a \sqrt { a } - 1} { a - \sqrt { a } } - \frac { a \sqrt { a } + 1} { a + \sqrt { a } }#?

1 Answer
George C.
Nov 23, 2017

#(asqrt(a)-1)/(a-sqrt(a))-(asqrt(a)+1)/(a+sqrt(a)) = 2#

Explanation:

Let #t=sqrt(a)#

Then:

#(asqrt(a)-1)/(a-sqrt(a))-(asqrt(a)+1)/(a+sqrt(a)) = (t^3-1)/(t^2-t)-(t^3+1)/(t^2+t)#

#color(white)((asqrt(a)-1)/(a-sqrt(a))-(asqrt(a)+1)/(a+sqrt(a))) = (color(red)(cancel(color(black)((t-1))))(t^2+t+1))/(t color(red)(cancel(color(black)((t-1)))))-(color(red)(cancel(color(black)((t+1))))(t^2-t+1))/(t color(red)(cancel(color(black)((t+1)))))#

#color(white)((asqrt(a)-1)/(a-sqrt(a))-(asqrt(a)+1)/(a+sqrt(a))) = (color(red)(cancel(color(black)(t^2)))+t+color(red)(cancel(color(black)(1))))/t-(color(red)(cancel(color(black)(t^2)))-t+color(red)(cancel(color(black)(1))))/t#

#color(white)((asqrt(a)-1)/(a-sqrt(a))-(asqrt(a)+1)/(a+sqrt(a))) = (2t)/t#

#color(white)((asqrt(a)-1)/(a-sqrt(a))-(asqrt(a)+1)/(a+sqrt(a))) = 2#

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