How to solve this?

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1 Answer
George C.
Aug 11, 2017

B. #(1-sqrt(x))/(1-x)#

Explanation:

I will use a couple of identities to help solve this:

Difference of squares:

#a^2-b^2 = (a-b)(a+b)#

Sum of cubes:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

#color(white)()#
We are given:

#((sqrt(x)-1)/(x-x^(1/2)+1) -: (x-1)/(x^(3/2)+1)) * 1/(sqrt(x)+1)#

Putting:

#t = sqrt(x) = x^(1/2)#

this becomes:

#((t-1)/(t^2-t+1) -: (t^2-1)/(t^3+1)) * 1/(t+1)#

#= ((t-1)/(t^2-t+1) * (t^3+1)/(t^2-1)) * 1/(t+1)#

#= ((color(red)(cancel(color(black)(t-1))))/(color(blue)(cancel(color(black)(t^2-t+1)))) * ((color(green)(cancel(color(black)(t+1))))(color(blue)(cancel(color(black)(t^2-t+1)))))/((color(red)(cancel(color(black)(t-1))))(color(green)(cancel(color(black)(t+1)))))) * 1/(t+1)#

#=1/(t+1)#

#=1/(sqrt(x)+1)#

We might like to stop at that, but it does not match any of the options given. Let us try rationalising the denominator by multiplying by the conjugate #(sqrt(x)-1)# ...

#=(sqrt(x)-1)/((sqrt(x)-1)(sqrt(x)+1))#

#=(sqrt(x)-1)/(x-1)#

#=(1-sqrt(x))/(1-x)#

i.e. B.

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