How do you rationalize the numerator and simplify #[(1/sqrtx)+9sqrtx]/(9x+1)#?

1 Answer
Apr 14, 2015

The result is #sqrtx/x#.

The reason is the following:
1st) You have to rationalize #1/sqrtx#. This is done by multiplying both numerator and denominator by #sqrtx#. By doing this, you obtain the following: #((1/sqrtx) + 9sqrtx)/(9x+1) = ((sqrtx/x)+9sqrtx)/(9x+1)#.
2nd) Now, you make "x" the common denominator of the numerator as follows:
#((sqrtx/x)+9sqrtx)/(9x+1) = ((sqrtx+9xsqrtx)/x)/(9x+1)#.
3rd) Now, you pass the intermediate "x" to the denominator:
#((sqrtx+9xsqrtx)/x)/(9x+1) = (sqrtx+9xsqrtx)/(x(9x+1))#.
4th) Now, you take common factor #sqrtx# from the numerator:
#(sqrtx+9xsqrtx)/(x(9x+1)) = (sqrtx(9x+1))/(x(9x+1)#.
5th) And, finally, you simplify the factor (9x+1) appearing both in the numerator and the denominator:
#(sqrtx(cancel(9x+1)))/(x(cancel(9x+1))) = sqrtx/x#.