How to solve this?
1 Answer
B.
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)#
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)/(x-1)#
#=(1-sqrt(x))/(1-x)#
i.e. B.