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

Apr 14, 2015

The result is $\frac{\sqrt{x}}{x}$.

The reason is the following:
1st) You have to rationalize $\frac{1}{\sqrt{x}}$. This is done by multiplying both numerator and denominator by $\sqrt{x}$. By doing this, you obtain the following: $\frac{\left(\frac{1}{\sqrt{x}}\right) + 9 \sqrt{x}}{9 x + 1} = \frac{\left(\frac{\sqrt{x}}{x}\right) + 9 \sqrt{x}}{9 x + 1}$.
2nd) Now, you make "x" the common denominator of the numerator as follows:
$\frac{\left(\frac{\sqrt{x}}{x}\right) + 9 \sqrt{x}}{9 x + 1} = \frac{\frac{\sqrt{x} + 9 x \sqrt{x}}{x}}{9 x + 1}$.
3rd) Now, you pass the intermediate "x" to the denominator:
$\frac{\frac{\sqrt{x} + 9 x \sqrt{x}}{x}}{9 x + 1} = \frac{\sqrt{x} + 9 x \sqrt{x}}{x \left(9 x + 1\right)}$.
4th) Now, you take common factor $\sqrt{x}$ 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:
$\frac{\sqrt{x} \left(\cancel{9 x + 1}\right)}{x \left(\cancel{9 x + 1}\right)} = \frac{\sqrt{x}}{x}$.