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