Lời giải:
$3-2x-\frac{1}{3}=7x-\frac{1}{4}$
$3-\frac{1}{3}+\frac{1}{4}=2x+7x$
$\frac{35}{12}=9x$
$x=\frac{35}{108}$
\(3-2x-\dfrac{1}{3}=7x-\dfrac{1}{4}\)
\(\Leftrightarrow3-\dfrac{1}{3}+\dfrac{1}{4}=7x-2x\)
\(\Leftrightarrow5x=\dfrac{35}{12}\)
\(\Leftrightarrow x=\dfrac{7}{12}\)