$a,\dfrac{2}{3}-\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{2}\left(2x+1\right)=5$

$\dfrac{2}{3}-\dfrac{1}{3}x-\dfrac{1}{2}-x+\dfrac{1}{2}=5$

$\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}x-x=5$

$\dfrac{2}{3}-\dfrac{1}{3}x-x=5$

$\dfrac{2}{3}-\dfrac{4}{3}x=5$

$\dfrac{4}{3}x=\dfrac{2}{3}-5$

$\dfrac{4}{3}x=-\dfrac{13}{3}$

$x=-\dfrac{13}{3}:\dfrac{4}{3}$

$x=-\dfrac{13}{4}$

Vậy=...

$b,\left(x+\dfrac{1}{2}\right)\left(\dfrac{3}{4}-x\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.$

Vậy=-

$c,\dfrac{2x-1}{-3+2}=0$

$\Rightarrow2x-1=0$

$\Rightarrow x=\dfrac{1}{2}$

Vậy=.