## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{3x-1}{3x}, \quad x>0$ or$\quad 1-\displaystyle \frac{1}{3x}, \quad x>0$
For $\sqrt{x}$ to be defined, we must have $x>0.$ In this case, none of the denominators yield zero. No extra exclusions, for now. Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, multiplying both the numerator and denominator with $3\sqrt{x}$ would get rid of these fractions. $\displaystyle \frac{\sqrt{x}-\frac{1}{3\sqrt{x}}}{\sqrt{x}}\times\frac{3\sqrt{x}}{3\sqrt{x}}=\frac{3x-1}{3x}, \quad x>0$ or$\quad 1-\displaystyle \frac{1}{3x}, \quad x>0$