%@P:exocorcp
%@Dif:3
Retrouve le nombre caché à la place de $\square$.
\[\frac{87}{60}=\frac12+\frac14+\frac13+\frac16+\frac1{\square}\]
%@Correction:
\[\Eqalign{
\frac{87}{60}&=\frac12+\frac14+\frac13+\frac16+\frac1{\square}\cr
\frac{87}{60}&=\frac24+\frac14+\frac26+\frac16+\frac1{\square}\cr
\frac{87}{60}&=\frac34+\frac36+\frac1{\square}\cr
\frac{87}{60}&=\frac34+\frac12+\frac1{\square}\cr
\frac{87}{60}&=\frac34+\frac24+\frac1{\square}\cr
\frac{87}{60}&=\frac54+\frac1{\square}\cr
\frac{87}{60}-\frac54&=\frac1{\square}\cr
\frac{87}{60}-\frac{75}{60}&=\frac1{\square}\cr
\frac{12}{60}&=\frac1{\square}\cr
\frac15&=\frac1{\square}\cr
5&=\square\cr
}\]