$3^2+3.(\dfrac{1}{2})^0-\dfrac{1}{4}-2.(-\dfrac{1}{2})^2-\dfrac{1}{2}$
$=9+3.1-\dfrac{1}{4}-2.\dfrac{1}{4}-\dfrac{1}{2}$
$=9+3-\dfrac{1}{4}-\dfrac{1}{2}-\dfrac{1}{2}$
$=12-\dfrac{1}{4}+(-\dfrac{1}{2}-\dfrac{1}{2})$
$=\dfrac{47}{4}-\dfrac{4}{4}=\dfrac{43}{4}$