Bài 3:
a: Ta có: \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow x-\dfrac{7}{6}=\dfrac{13}{12}\)
hay \(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13+14}{12}=\dfrac{27}{12}=\dfrac{9}{4}\)
c: Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
hay \(x=\dfrac{7}{2}\)