\(\dfrac{3}{4}-\sqrt[]{\dfrac{3}{12}+\dfrac{\left(\sqrt[]{3^2}\right)}{4}}\)
\(=\dfrac{3}{4}-\sqrt[]{\dfrac{1}{4}+\dfrac{3}{4}}\)
\(=\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{3}{4}\)
\(=1\)
\(\dfrac{3}{4}-\sqrt{\dfrac{1}{4}}+\dfrac{3}{4}=\dfrac{6}{4}-\dfrac{1}{2}=1\)