Answer
$\displaystyle\frac{3x+2}{\sqrt{x}\sqrt{3x+4}}$
Work Step by Step
Factor out $x$ and $3x+4$: $\displaystyle \frac{1}{2}x^{-1/2}(3x+4)^{1/2}+\frac{3}{2}x^{1/2}(3x+4)^{-1/2}=\frac{1}{2}x^{-1/2}(3x+4)^{-1/2}[(3x+4)+3x]=\frac{1}{2}x^{-1/2}(3x+4)^{-1/2}(6x+4) =x^{-1/2}(3x+4)^{-1/2}(3x+2)=\frac{3x+2}{\sqrt{x}\sqrt{3x+4}}$
