Avoid using \startformula \stopformula=C2=A0in itemize. Us= e \dfrac to get display=C2=A0style fraction.

\define= enumeration
=C2=A0 [exercici]
=C2=A0 [alternative=3Dserried, before= =3D{\blank[medium]\testpage[1]}, text=3D{Exercici}, stopper=3D{.\space}, wi= dth=3Dfit, headstyle=3D\ss, distance=3D0.25em, width=3Dfit, headcolor=3Dexe= rcici_color, title=3Dyes, titlestyle=3D\ss, titledistance=3D0em, left=3D{\b= group\bf}, right=3D{\egroup}, after=3D{\blank[big]}, prefix=3Dyes, prefixse= gments=3Dsection, way=3Dsection]
\starttext

\section{Operacions a= mb fraccions}

\subject{Preguntes}

\startexercici[exercici:ope= racions-amb-fraccions-1]{suma i resta} Opereu:
\startitemize[a, columns,= three]
\item\m{\dfrac{2}{7} + \dfrac{6}{7} - \dfrac{1}{7}}
\item\m{\= dfrac{5}{8} + \dfrac{9}{8}}
\item\m{\dfrac{1}{4} + 5 - \dfrac{1}{4}}
= \item\m{\dfrac{3}{5} - \dfrac{2}{5} + \dfrac{11}{5}}
\item\m{\dfrac{2}{3= } + \dfrac{1}{4}}
\item\m{5 + \dfrac{4}{5}}
\item\m{\dfrac{2}{3} + \d= frac{5}{8} - \dfrac{1}{9}}
\item\m{10 - \dfrac{1}{6} + \dfrac{6}{8}}
= \item\m{\dfrac{3}{10} + \dfrac{7}{12}}
\stopitemize
\stopexercici
=
\startexercici[exercici:operacions-amb-fraccions-2]{multiplicaci=C3=B3 = i divisi=C3=B3} Opereu:
\startitemize[a, columns, three]
\item\m{3 \c= dot \dfrac{5}{3}}
\item\m{\dfrac{3}{4} \cdot \dfrac{2}{5}}
\item\m{\d= frac{1}{3} \cdot \dfrac{3}{4} \div \dfrac{1}{5}}
%\column
\item\m{5 \= div \dfrac{7}{10} \cdot 7 \div \dfrac{2}{5}}
\item\m{\dfrac{1}{2} \cdot = \dfrac{1}{2}}
\item\m{\dfrac{3}{2} \div \dfrac{2}{3} \div \dfrac{5}{4}}<= br>%\column
\item\m{\dfrac{2}{3} \colon \dfrac{2}{5}}
\item\m{\dfrac{= 3}{4} \colon \dfrac{1}{2} \cdot 6}
\item\m{\dfrac{2}{3} \colon \dfrac{3}= {4} \cdot \dfrac{1}{2}}
\stopitemize
\stopexercici

\startexerc= ici[exercici:operacions-amb-fraccions-3]{operacions combinades}
\startit= emize[a, columns, three]
\item\m{\dfrac{3}{4} + \dfrac{2}{5} \cdot \dfra= c{1}{2}}
\item\m{\dfrac{2}{5} \div \dfrac{3}{5} - \dfrac{1}{4}}
\item= \m{\left( 2 + \dfrac{3}{5} \right) \div \dfrac{4}{3}}
\item\m{\dfrac{2}{= 3} - \dfrac{5}{6}\cdot 8}
\item\m{4 + \dfrac{3}{5} - \left(2 - \dfrac{1}= {2}\right)}
\item\m{4 \cdot \left(\dfrac{5}{6} - \dfrac{1}{4} + \dfrac{4= }{9}\right)}
\item\m{\dfrac{5}{6} - \left(\dfrac{1}{5} + \dfrac{1}{3}\ri= ght)}
\item\m{\dfrac{5}{6} \div \left( \dfrac{2}{5} - \dfrac{1}{10}\righ= t)}
\item\m{4 \cdot \dfrac{2}{3} \cdot \left(\dfrac{2}{4} \div \dfrac{8}= {10}\right)}
\item\m{\dfrac{3}{4} + \dfrac{1}{2} \cdot \dfrac{4}{10}}\item\m{\dfrac{10}{12} - \left(\dfrac{1}{10} + \dfrac{3}{8}\right)}
\it= em\m{\left(\dfrac{1}{2} + 2\right) \cdot \left(2 - \dfrac{1}{4}\right)}
= \item\m{\left(\dfrac{3}{4} + \dfrac{1}{4}\right) \div \left(\dfrac{5}{6} - = \dfrac{1}{3}\right)}
\item\m{5 + 10 - \dfrac{1}{10} + 25}
\item\m{\df= rac{1}{3} + \dfrac{1}{2} - \dfrac{2}{5}}
\item\m{\left(\dfrac{1}{2} + \d= frac{2}{10}\right) \cdot \dfrac{2}{6}}
\item\m{5 + \dfrac{1}{3} \cdot \d= frac{5}{3}}
\item\m{\left( 5 + \dfrac{1}{3} \right) \cdot \dfrac{5}{3}}<= br>\stopitemize
\stopexercici
\stoptext

Xavie= r B. via ntg-context <ntg-context@= ntg.nl> =E6=96=BC 2022=E5=B9=B42=E6=9C=8825=E6=97=A5 =E9=80=B1=E4=BA= =94 =E4=B8=8A=E5=8D=886:10=E5=AF=AB=E9=81=93=EF=BC=9A

Even if I change
\startexercici[exercici:operacions-amb-fraccions-2]{multiplicaci=C3=B3 i di= visi=C3=B3} Opereu:
\startitemize[a, columns, three]

for

\startexercici[exercici:operacions-amb-fraccions-2]{multiplicaci=C3=B3 i di= visi=C3=B3} Opereu:
\startitemize[a, columns]

the 'c' jumps.

Isn't ConTeXt capable of handle that?
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