En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Sujet de brevetFraction/Exercices/Sujet de brevet », n'a pu être restituée correctement ci-dessus.
Cet exercice est tombé au brevet des collèges (2000).
Calculer et donner le résultat sous la forme d'une fraction la plus simple possible :
A = 4 3 − 1 3 × ( 3 + 1 2 ) {\displaystyle A={\frac {4}{3}}-{\frac {1}{3}}\times \left(3+{\frac {1}{2}}\right)}
Solution
A = 4 3 − 1 3 × ( 3 + 1 2 ) {\displaystyle {\frac {4}{3}}-{\frac {1}{3}}\times \left(3+{\frac {1}{2}}\right)}
= 4 3 − 1 3 × ( 6 2 + 1 2 ) {\displaystyle {\frac {4}{3}}-{\frac {1}{3}}\times \left({\frac {6}{2}}+{\frac {1}{2}}\right)}
= 4 3 − 1 3 × ( 6 + 1 2 ) {\displaystyle {\frac {4}{3}}-{\frac {1}{3}}\times \left({\frac {6+1}{2}}\right)}
= 4 3 − 1 3 × 7 2 {\displaystyle {\frac {4}{3}}-{\frac {1}{3}}\times {\frac {7}{2}}}
= 4 3 − 7 6 {\displaystyle {\frac {4}{3}}-{\frac {7}{6}}}
= 8 6 − 7 6 {\displaystyle {\frac {8}{6}}-{\frac {7}{6}}}
= 1 6 {\displaystyle {\frac {1}{6}}}
Cet exercice est tombé au brevet des collèges (2001).
Donner ces résultats sous forme de fractions irréductibles :
A = ( 5 7 ) 2 − 2 7 {\displaystyle A=\left({\frac {5}{7}}\right)^{2}-{\frac {2}{7}}}
Solution
A = ( 5 7 ) 2 − 2 7 {\displaystyle A=\left({\frac {5}{7}}\right)^{2}-{\frac {2}{7}}}
= ( 25 49 ) − 2 7 {\displaystyle \left({\frac {25}{49}}\right)-{\frac {2}{7}}}
= ( 25 49 ) − 2 × 7 7 × 7 {\displaystyle \left({\frac {25}{49}}\right)-{\frac {2\times 7}{7\times 7}}}
= 25 49 − 14 49 {\displaystyle {\frac {25}{49}}-{\frac {14}{49}}}
= 25 − 14 49 {\displaystyle {\frac {25-14}{49}}}
= 11 49 {\displaystyle {\frac {11}{49}}}
B = 1 9 + 1 12 {\displaystyle B={\frac {1}{9}}+{\frac {1}{12}}}
Solution
B = 1 9 + 1 12 {\displaystyle B={\frac {1}{9}}+{\frac {1}{12}}}
= 1 × 12 9 × 12 + 1 × 9 12 × 9 {\displaystyle ={\frac {1\times 12}{9\times 12}}+{\frac {1\times 9}{12\times 9}}}
= 12 108 + 9 108 {\displaystyle ={\frac {12}{108}}+{\frac {9}{108}}}
= 12 + 9 108 {\displaystyle ={\frac {12+9}{108}}}
= 7 ∗ 3 36 ∗ 3 {\displaystyle ={\frac {7*3}{36*3}}}
= 7 36 {\displaystyle ={\frac {7}{36}}}
En électricité, pour calculer des valeurs de résistances, on utilise la formule: 1 R = 1 R 1 + 1 R 2 {\displaystyle {\frac {1}{R}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}}
Sachant que R 1 {\displaystyle R_{1}} R 2 {\displaystyle R_{2}}
Solution
1 R = 1 R 1 + 1 R 2 {\displaystyle {\frac {1}{R}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}}
= 1 R = 1 9 + 1 12 {\displaystyle {\frac {1}{R}}={\frac {1}{9}}+{\frac {1}{12}}}
= 1 R = 21 108 {\displaystyle {\frac {1}{R}}={\frac {21}{108}}}
R = 36 7 o h m s {\displaystyle R={\frac {36}{7}}ohms}
Cet exercice est tombé au brevet des collèges (2000).
Mettre sous la forme de fraction irréductible
11 3 − 7 25 6 {\displaystyle {\frac {{\frac {11}{3}}-7}{\frac {25}{6}}}}
Solution
11 3 − 7 25 6 {\displaystyle {\frac {{\frac {11}{3}}-7}{\frac {25}{6}}}}
= 11 × 2 3 × 2 − 7 × 6 6 25 6 {\displaystyle {\frac {{\frac {11\times 2}{3\times 2}}-{\frac {7\times 6}{6}}}{\frac {25}{6}}}}
= 22 6 − 42 6 25 6 {\displaystyle {\frac {{\frac {22}{6}}-{\frac {42}{6}}}{\frac {25}{6}}}}
= 22 − 42 25 {\displaystyle {\frac {22-42}{25}}}
= − 20 25 {\displaystyle -{\frac {20}{25}}}
= − 4 5 {\displaystyle -{\frac {4}{5}}}
Mettre sous la forme de fraction irréductible
3 7 − 2 5 × 15 4 {\displaystyle {\frac {3}{7}}-{\frac {2}{5}}\times {\frac {15}{4}}}
Solution
3 7 − 2 5 × 15 4 {\displaystyle {\frac {3}{7}}-{\frac {2}{5}}\times {\frac {15}{4}}}
= 3 7 − 2 × 15 5 × 4 {\displaystyle {\frac {3}{7}}-{\frac {2\times 15}{5\times 4}}}
= 3 × 20 7 × 20 − 2 × 15 × 7 5 × 4 × 7 {\displaystyle {\frac {3\times 20}{7\times 20}}-{\frac {2\times 15\times 7}{5\times 4\times 7}}}
= 60 140 − 210 140 {\displaystyle {\frac {60}{140}}-{\frac {210}{140}}}
= 60 − 210 140 {\displaystyle {\frac {60-210}{140}}}
= − 150 140 {\displaystyle -{\frac {150}{140}}}
= − 15 14 {\displaystyle -{\frac {15}{14}}}
Cet exercice est tombé au brevet des collèges (2000).
calculer A = ( 1 3 − 1 5 ) ÷ 2 5 {\displaystyle A=\left({\frac {1}{3}}-{\frac {1}{5}}\right)\div {\frac {2}{5}}}
Solution
A = ( 1 3 − 1 5 ) ÷ 2 5 {\displaystyle A=\left({\frac {1}{3}}-{\frac {1}{5}}\right)\div {\frac {2}{5}}}
=( 1 × 5 3 × 5 − 1 × 3 5 × 3 ) ÷ 2 5 {\displaystyle \left({\frac {1\times 5}{3\times 5}}-{\frac {1\times 3}{5\times 3}}\right)\div {\frac {2}{5}}}
=( 5 15 − 3 15 ) ÷ 2 5 {\displaystyle \left({\frac {5}{15}}-{\frac {3}{15}}\right)\div {\frac {2}{5}}}
=5 − 3 15 ÷ 2 5 {\displaystyle {\frac {5-3}{15}}\div {\frac {2}{5}}}
=2 15 2 5 {\displaystyle {\frac {\frac {2}{15}}{\frac {2}{5}}}}
=2 15 × 5 2 {\displaystyle {\frac {2}{15}}\times {\frac {5}{2}}}
=2 × 5 15 × 2 {\displaystyle {\frac {2\times 5}{15\times 2}}}
= 10 30 {\displaystyle {\frac {10}{30}}}
=1 3 {\displaystyle {\frac {1}{3}}}
Calculer ( ( − 4 + 3 ) × 2 7 ) ÷ 3 14 {\displaystyle \left((-4+3)\times {\frac {2}{7}}\right)\div {\frac {3}{14}}}
Solution
( ( − 4 + 3 ) × 2 7 ) ÷ 3 14 {\displaystyle \left((-4+3)\times {\frac {2}{7}}\right)\div {\frac {3}{14}}}
= ( − 1 × 2 7 ) ÷ 3 14 {\displaystyle \left(-1\times {\frac {2}{7}}\right)\div {\frac {3}{14}}}
= ( − 2 7 ) ÷ 3 14 {\displaystyle \left(-{\frac {2}{7}}\right)\div {\frac {3}{14}}}
= − 2 7 × 14 3 {\displaystyle -{\frac {2}{7}}\times {\frac {14}{3}}}
= − 2 × 14 7 × 3 {\displaystyle -{\frac {2\times 14}{7\times 3}}}
= − 2 × 2 × 7 7 × 3 {\displaystyle -{\frac {2\times 2\times 7}{7\times 3}}}
= − 2 × 2 3 {\displaystyle -{\frac {2\times 2}{3}}}
= − 4 3 {\displaystyle -{\frac {4}{3}}}
Calculer 4 − ( 2 − 5 ) 2 4 + 5 {\displaystyle {\frac {4-\left(2-5\right)^{2}}{4+5}}}
Solution
4 − ( 2 − 5 ) 2 4 + 5 {\displaystyle {\frac {4-\left(2-5\right)^{2}}{4+5}}}
= 4 − ( − 3 ) 2 9 {\displaystyle {\frac {4-\left(-3\right)^{2}}{9}}}
= 4 − 9 9 {\displaystyle {\frac {4-9}{9}}}
= − 5 9 {\displaystyle {\frac {-5}{9}}}
Calculer 3 − 3 ÷ 9 2 {\displaystyle 3-3\div {\frac {9}{2}}}
Solution
3 − 3 ÷ 9 2 {\displaystyle 3-3\div {\frac {9}{2}}}
= 3 − 3 × 2 9 {\displaystyle 3-3\times {\frac {2}{9}}}
= 3 − 3 × 2 9 {\displaystyle 3-{\frac {3\times 2}{9}}}
= 3 − 2 3 {\displaystyle 3-{\frac {2}{3}}}
= 9 3 − 2 3 {\displaystyle {\frac {9}{3}}-{\frac {2}{3}}}
= 9 − 2 3 {\displaystyle {\frac {9-2}{3}}}
= 7 3 {\displaystyle {\frac {7}{3}}}
Cet exercice est tombé au brevet des collèges (1996).
Calculer A = 7 18 × 2 7 − ( 5 3 − 1 ) 2 {\displaystyle A={\frac {7}{18}}\times {\frac {2}{7}}-\left({\frac {5}{3}}-1\right)^{2}}
Solution
A = 7 18 × 2 7 − ( 5 3 − 1 ) 2 {\displaystyle A={\frac {7}{18}}\times {\frac {2}{7}}-\left({\frac {5}{3}}-1\right)^{2}}
= 7 × 2 18 × 7 − ( 5 3 − 3 3 ) 2 {\displaystyle {\frac {7\times 2}{18\times 7}}-\left({\frac {5}{3}}-{\frac {3}{3}}\right)^{2}}
= 2 18 − ( 5 − 3 3 ) 2 {\displaystyle {\frac {2}{18}}-\left({\frac {5-3}{3}}\right)^{2}}
= 1 9 − ( 2 3 ) 2 {\displaystyle {\frac {1}{9}}-\left({\frac {2}{3}}\right)^{2}}
= 1 9 − ( 4 9 ) {\displaystyle {\frac {1}{9}}-\left({\frac {4}{9}}\right)}
= − 3 9 {\displaystyle -{\frac {3}{9}}}
= − 1 3 {\displaystyle -{\frac {1}{3}}}
A = 5 7 − 2 7 × 4 3 {\displaystyle A={\frac {5}{7}}-{\frac {2}{7}}\times {\frac {4}{3}}}
Solution
A = 5 7 − 2 7 × 4 3 {\displaystyle A={\frac {5}{7}}-{\frac {2}{7}}\times {\frac {4}{3}}}
= 5 × 3 7 × 3 − 2 × 4 7 × 3 {\displaystyle {\frac {5\times 3}{7\times 3}}-{\frac {2\times 4}{7\times 3}}}
= 15 21 − 8 21 {\displaystyle {\frac {15}{21}}-{\frac {8}{21}}}
= 15 − 8 21 {\displaystyle {\frac {15-8}{21}}}
= 7 21 {\displaystyle {\frac {7}{21}}}
= 1 3 {\displaystyle {\frac {1}{3}}}
B = ( 1 2 − 1 3 ) × 3 2 + 1 {\displaystyle B=\left({\frac {1}{2}}-{\frac {1}{3}}\right)\times {\frac {3}{2}}+1}
Solution
B = ( 1 2 − 1 3 ) × 3 2 + 1 {\displaystyle B=\left({\frac {1}{2}}-{\frac {1}{3}}\right)\times {\frac {3}{2}}+1}
= ( 3 6 − 2 6 ) × 3 2 + 1 {\displaystyle \left({\frac {3}{6}}-{\frac {2}{6}}\right)\times {\frac {3}{2}}+1}
= ( 1 6 ) × 3 2 + 1 {\displaystyle \left({\frac {1}{6}}\right)\times {\frac {3}{2}}+1}
= 3 12 + 1 {\displaystyle {\frac {3}{12}}+1}
= 15 12 {\displaystyle {\frac {15}{12}}}
= 5 4 {\displaystyle {\frac {5}{4}}}
Cet exercice est tombé au brevet des collèges (1996).
A = 3 2 − 5 2 × 3 10 {\displaystyle A={\frac {3}{2}}-{\frac {5}{2}}\times {\frac {3}{10}}}
Solution
A = 3 2 − 5 2 × 3 10 {\displaystyle A={\frac {3}{2}}-{\frac {5}{2}}\times {\frac {3}{10}}}
= 3 2 − 15 20 {\displaystyle {\frac {3}{2}}-{\frac {15}{20}}}
= 30 20 − 15 20 {\displaystyle {\frac {30}{20}}-{\frac {15}{20}}}
= 15 20 {\displaystyle {\frac {15}{20}}}
= 3 4 {\displaystyle {\frac {3}{4}}}
B = ( 3 5 ) 2 ÷ 9 20 {\displaystyle B=\left({\frac {3}{5}}\right)^{2}\div {\frac {9}{20}}}
Solution
B = ( 3 5 ) 2 ÷ 9 20 {\displaystyle B=\left({\frac {3}{5}}\right)^{2}\div {\frac {9}{20}}}
= ( 9 25 ) × 20 9 {\displaystyle \left({\frac {9}{25}}\right)\times {\frac {20}{9}}}
= 20 25 {\displaystyle {\frac {20}{25}}}
= 4 5 {\displaystyle {\frac {4}{5}}}
Cet exercice est tombé au brevet des collèges (1996).
On pose A = 4 − 3 4 ( 1 3 − 1 6 ) {\displaystyle A=4-{\frac {3}{4}}\left({\frac {1}{3}}-{\frac {1}{6}}\right)}
En faisant apparaître les étapes de calcul, donner une écriture fractionnaire et une écriture décimale du nombre A.
Solution
A = 4 − 3 4 ( 1 3 − 1 6 ) {\displaystyle A=4-{\frac {3}{4}}\left({\frac {1}{3}}-{\frac {1}{6}}\right)}
= 4 − 3 4 ( 2 6 − 1 6 ) {\displaystyle =4-{\frac {3}{4}}\left({\frac {2}{6}}-{\frac {1}{6}}\right)}
= 4 − 3 4 ( 1 6 ) {\displaystyle =4-{\frac {3}{4}}\left({\frac {1}{6}}\right)}
= 4 − 3 24 {\displaystyle =4-{\frac {3}{24}}}
= 4 − 1 8 {\displaystyle =4-{\frac {1}{8}}}
= 32 8 − 1 8 {\displaystyle ={\frac {32}{8}}-{\frac {1}{8}}}
= 31 8 {\displaystyle ={\frac {31}{8}}}
= 3 , 875 {\displaystyle =3,875}
On retire une petite partie de 4, donc on sait que le résultat est relativement proche de 4 au final.
Donner la valeur exacte la plus simple possible en indiquant le détail des calculs :
34 5 ÷ ( 4 5 − 3 8 ) {\displaystyle {\frac {34}{5}}\div \left({\frac {4}{5}}-{\frac {3}{8}}\right)}
Solution
= 34 5 ÷ ( 4 5 × 8 8 − 3 8 × 5 5 ) {\displaystyle ={\frac {34}{5}}\div \left({\frac {4}{5}}\times {\frac {8}{8}}-{\frac {3}{8}}\times {\frac {5}{5}}\right)}
= 34 5 ÷ ( 4 5 × 8 8 − 3 8 × 5 5 ) {\displaystyle ={\frac {34}{5}}\div \left({\frac {4}{5}}\times {\frac {8}{8}}-{\frac {3}{8}}\times {\frac {5}{5}}\right)}
= 34 5 ÷ ( 32 − 15 5 × 8 ) {\displaystyle ={\frac {34}{5}}\div \left({\frac {32-15}{5\times 8}}\right)}
= 34 5 ÷ ( 17 5 × 8 ) {\displaystyle ={\frac {34}{5}}\div \left({\frac {17}{5\times 8}}\right)}
= 34 5 × ( 5 × 8 17 ) {\displaystyle ={\frac {34}{5}}\times \left({\frac {5\times 8}{17}}\right)}
= 17 × 2 5 × ( 5 × 8 17 ) {\displaystyle ={\frac {17\times 2}{5}}\times \left({\frac {5\times 8}{17}}\right)}
= 2 × 8 = 16 {\displaystyle =2\times 8=16}
24 × 10 − 2 × 3 , 5 × 10 5 8 × 10 − 1 × 21 × 10 4 {\displaystyle {\frac {24\times 10^{-2}\times 3,5\times 10^{5}}{8\times 10^{-1}\times 21\times 10^{4}}}}
Solution
= 24 × 3 , 5 8 × 21 × 10 − 2 × 10 5 10 − 1 × 10 4 {\displaystyle ={\frac {24\times 3,5}{8\times 21}}\times {\frac {10^{-2}\times 10^{5}}{10^{-1}\times 10^{4}}}}
= 8 × 3 × 3 , 5 8 × 3 × 7 × 10 3 10 3 {\displaystyle ={\frac {8\times 3\times 3,5}{8\times 3\times 7}}\times {\frac {10^{3}}{10^{3}}}}
= 3 , 5 7 {\displaystyle ={\frac {3,5}{7}}}
= 1 2 {\displaystyle ={\frac {1}{2}}}
Cet exercice est tombé au brevet des collèges (1995).
On considère les nombres :
A = 6 7 − 4 7 × 5 2 {\displaystyle A={\frac {6}{7}}-{\frac {4}{7}}\times {\frac {5}{2}}} B = 3 4 − 4 3 4 + 1 3 {\displaystyle B={\frac {{\frac {3}{4}}-4}{{\frac {3}{4}}+{\frac {1}{3}}}}} C = 3 2 × 2 − 125 × 10 − 1 {\displaystyle C=3^{2}\times 2-125\times 10^{-1}} En précisant les différentes étapes des calculs
a) Écrire A sous la forme la plus simple possible et sans utiliser de valeur approchée
Solution
A = 6 7 − 4 7 × 5 2 {\displaystyle A={\frac {6}{7}}-{\frac {4}{7}}\times {\frac {5}{2}}}
A = 6 7 − 4 × 5 7 × 2 {\displaystyle A={\frac {6}{7}}-{\frac {4\times 5}{7\times 2}}}
A = 6 7 − 20 14 {\displaystyle A={\frac {6}{7}}-{\frac {20}{14}}}
A = 12 14 − 20 14 {\displaystyle A={\frac {12}{14}}-{\frac {20}{14}}}
A = 12 − 20 14 {\displaystyle A={\frac {12-20}{14}}}
A = − 8 14 {\displaystyle A={\frac {-8}{14}}}
A = − 4 7 {\displaystyle A={\frac {-4}{7}}}
b) Écrire B sous la forme d'une nombre entier relatif
c) Écrire C sous la forme d'un nombre décimal
Cet exercice est tombé au brevet des collèges (1995).
Sachant que a = 2 3 {\displaystyle a={\frac {2}{3}}} b = − 1 4 {\displaystyle b=-{\frac {1}{4}}} c = 2 5 {\displaystyle c={\frac {2}{5}}} d = − 1 2 {\displaystyle d=-{\frac {1}{2}}}
a) Calculer A = a b + c d {\displaystyle A=ab+cd}
Solution
A = a b + c d {\displaystyle A=ab+cd}
A = 2 3 × − 1 4 + 2 5 × − 1 2 {\displaystyle A={\frac {2}{3}}\times {\frac {-1}{4}}+{\frac {2}{5}}\times {\frac {-1}{2}}}
A = − 2 12 + − 2 10 {\displaystyle A={\frac {-2}{12}}+{\frac {-2}{10}}}
A = − 10 60 + − 12 60 {\displaystyle A={\frac {-10}{60}}+{\frac {-12}{60}}}
A = − 22 60 {\displaystyle A={\frac {-22}{60}}}
A = − 11 30 {\displaystyle A={\frac {-11}{30}}}
b) Calculer B = a + d b + c {\displaystyle B={\frac {a+d}{b+c}}}
Solution
B = a + d b + c {\displaystyle B={\frac {a+d}{b+c}}}
B = 2 3 − 1 2 − 1 4 + 2 5 {\displaystyle B={\frac {{\frac {2}{3}}-{\frac {1}{2}}}{-{\frac {1}{4}}+{\frac {2}{5}}}}}
B = 4 6 − 3 6 − 5 20 + 8 20 {\displaystyle B={\frac {{\frac {4}{6}}-{\frac {3}{6}}}{{\frac {-5}{20}}+{\frac {8}{20}}}}}
B = 1 6 3 20 {\displaystyle B={\frac {\frac {1}{6}}{\frac {3}{20}}}}
B = 1 × 20 6 × 3 {\displaystyle B={\frac {1\times 20}{6\times 3}}}
B = 20 18 {\displaystyle B={\frac {20}{18}}}
B = 10 9 {\displaystyle B={\frac {10}{9}}}
Cet exercice est tombé au brevet des collèges (1995).
Calculer et donner chaque résultat sous la forme d'une fraction aussi simple que possible
A = 3 4 − 5 6 × 3 2 {\displaystyle A={\frac {3}{4}}-{\frac {5}{6}}\times {\frac {3}{2}}}
Solution
A = 3 4 − 5 6 × 3 2 {\displaystyle A={\frac {3}{4}}-{\frac {5}{6}}\times {\frac {3}{2}}}
A = 3 4 − 5 × 3 6 × 2 {\displaystyle A={\frac {3}{4}}-{\frac {5\times 3}{6\times 2}}}
A = 3 4 − 15 12 {\displaystyle A={\frac {3}{4}}-{\frac {15}{12}}}
A = 9 12 − 15 12 {\displaystyle A={\frac {9}{12}}-{\frac {15}{12}}}
A = 9 − 15 12 {\displaystyle A={\frac {9-15}{12}}}
A = − 6 12 {\displaystyle A={\frac {-6}{12}}}
A = − 1 2 {\displaystyle A=-{\frac {1}{2}}}
Cet exercice est tombé au brevet des collèges (2002).
Calculer :
4 5 − 2 × 6 5 {\displaystyle {\frac {4}{5}}-2\times {\frac {6}{5}}}
Solution
4 5 − 2 × 6 5 {\displaystyle {\frac {4}{5}}-2\times {\frac {6}{5}}}
4 5 − 12 5 {\displaystyle {\frac {4}{5}}-{\frac {12}{5}}}
4 − 12 5 {\displaystyle {\frac {4-12}{5}}}
− 8 5 {\displaystyle {\frac {-8}{5}}}
Cet exercice est tombé au brevet des collèges (1996).
Mettre sous la forme la plus simple le nombre 4 5 × 2 3 + 1 2 − 1 30 {\displaystyle {\frac {4}{5}}\times {\frac {2}{3}}+{\frac {1}{2}}-{\frac {1}{30}}}
Solution
4 5 × 2 3 + 1 2 − 1 30 {\displaystyle {\frac {4}{5}}\times {\frac {2}{3}}+{\frac {1}{2}}-{\frac {1}{30}}}
= 4 × 2 5 × 3 + 1 2 − 1 30 {\displaystyle ={\frac {4\times 2}{5\times 3}}+{\frac {1}{2}}-{\frac {1}{30}}}
= 8 15 + 1 2 − 1 30 {\displaystyle ={\frac {8}{15}}+{\frac {1}{2}}-{\frac {1}{30}}}
= 16 30 + 1 2 − 1 30 {\displaystyle ={\frac {16}{30}}+{\frac {1}{2}}-{\frac {1}{30}}}
= 16 30 + 15 30 − 1 30 {\displaystyle ={\frac {16}{30}}+{\frac {15}{30}}-{\frac {1}{30}}}
= 16 + 15 − 1 30 {\displaystyle ={\frac {16+15-1}{30}}}
= 30 30 {\displaystyle ={\frac {30}{30}}}
= 1 {\displaystyle =1}
Cet exercice est tombé au brevet des collèges (1996).
Écrire le résultat sous la forme la plus simple
A = 5 7 − 2 7 × 4 3 {\displaystyle A={\frac {5}{7}}-{\frac {2}{7}}\times {\frac {4}{3}}}
Solution
A = 5 7 − 2 7 × 4 3 {\displaystyle A={\frac {5}{7}}-{\frac {2}{7}}\times {\frac {4}{3}}}
A = 5 7 − 2 × 4 7 × 3 {\displaystyle A={\frac {5}{7}}-{\frac {2\times 4}{7\times 3}}}
A = 5 7 − 8 21 {\displaystyle A={\frac {5}{7}}-{\frac {8}{21}}}
A = 15 21 − 8 21 {\displaystyle A={\frac {15}{21}}-{\frac {8}{21}}}
A = 15 − 8 21 {\displaystyle A={\frac {15-8}{21}}}
A = 7 21 = 1 3 {\displaystyle A={\frac {7}{21}}={\frac {1}{3}}}
B = ( 1 2 − 1 3 ) ÷ 2 3 + 1 {\displaystyle B=\left({\frac {1}{2}}-{\frac {1}{3}}\right)\div {\frac {2}{3}}+1}
Solution
B = ( 1 2 − 1 3 ) ÷ 2 3 + 1 {\displaystyle B=\left({\frac {1}{2}}-{\frac {1}{3}}\right)\div {\frac {2}{3}}+1}
B = ( 3 6 − 2 6 ) ÷ 2 3 + 1 {\displaystyle B=\left({\frac {3}{6}}-{\frac {2}{6}}\right)\div {\frac {2}{3}}+1}
B = ( 3 − 2 6 ) ÷ 2 3 + 1 {\displaystyle B=\left({\frac {3-2}{6}}\right)\div {\frac {2}{3}}+1}
B = ( 1 6 ) ÷ 2 3 + 1 {\displaystyle B=\left({\frac {1}{6}}\right)\div {\frac {2}{3}}+1}
B = 1 6 ÷ 2 3 + 1 {\displaystyle B={\frac {1}{6}}\div {\frac {2}{3}}+1}
B = 1 6 × 3 2 + 1 {\displaystyle B={\frac {1}{6}}\times {\frac {3}{2}}+1}
B = 1 × 3 6 × 2 + 1 {\displaystyle B={\frac {1\times 3}{6\times 2}}+1}
B = 3 12 + 1 {\displaystyle B={\frac {3}{12}}+1}
B = 3 12 + 12 12 {\displaystyle B={\frac {3}{12}}+{\frac {12}{12}}}
B = 3 + 12 12 {\displaystyle B={\frac {3+12}{12}}}
B = 15 12 = 5 4 {\displaystyle B={\frac {15}{12}}={\frac {5}{4}}}
Cet exercice est tombé au brevet des collèges (1996).
Calculer A et B. On donnera les résultats sous la forme la plus simple possible
A = 1 3 × 4 + 7 6 {\displaystyle A={\frac {1}{3}}\times 4+{\frac {7}{6}}}
Solution
A = 1 3 × 4 + 7 6 {\displaystyle A={\frac {1}{3}}\times 4+{\frac {7}{6}}}
A = 1 × 4 3 + 7 6 {\displaystyle A={\frac {1\times 4}{3}}+{\frac {7}{6}}}
A = 4 3 + 7 6 {\displaystyle A={\frac {4}{3}}+{\frac {7}{6}}}
A = 8 6 + 7 6 {\displaystyle A={\frac {8}{6}}+{\frac {7}{6}}}
A = 8 + 7 6 {\displaystyle A={\frac {8+7}{6}}}
A = 15 6 {\displaystyle A={\frac {15}{6}}}
B = 2 5 × 3 4 − 2 1 − 2 7 {\displaystyle B={\frac {2}{5}}\times {\frac {3}{4}}-{\frac {2}{1-{\frac {2}{7}}}}}
Solution
B = 2 5 × 3 4 − 2 1 − 2 7 {\displaystyle B={\frac {2}{5}}\times {\frac {3}{4}}-{\frac {2}{1-{\frac {2}{7}}}}}
B = 2 × 3 5 × 4 − 2 1 − 2 7 {\displaystyle B={\frac {2\times 3}{5\times 4}}-{\frac {2}{1-{\frac {2}{7}}}}}
B = 6 20 − 2 1 − 2 7 {\displaystyle B={\frac {6}{20}}-{\frac {2}{1-{\frac {2}{7}}}}}
B = 6 20 − 2 7 7 − 2 7 {\displaystyle B={\frac {6}{20}}-{\frac {2}{{\frac {7}{7}}-{\frac {2}{7}}}}}
B = 6 20 − 2 7 − 2 7 {\displaystyle B={\frac {6}{20}}-{\frac {2}{\frac {7-2}{7}}}}
B = 6 20 − 2 5 7 {\displaystyle B={\frac {6}{20}}-{\frac {2}{\frac {5}{7}}}}
B = 6 20 − 2 × 7 5 {\displaystyle B={\frac {6}{20}}-2\times {\frac {7}{5}}}
B = 6 20 − 2 × 7 5 {\displaystyle B={\frac {6}{20}}-{\frac {2\times 7}{5}}}
B = 6 20 − 14 5 {\displaystyle B={\frac {6}{20}}-{\frac {14}{5}}}
B = 3 10 − 28 10 {\displaystyle B={\frac {3}{10}}-{\frac {28}{10}}}
B = 3 − 28 10 {\displaystyle B={\frac {3-28}{10}}}
B = − 25 10 {\displaystyle B={\frac {-25}{10}}}
B = − 5 2 {\displaystyle B={\frac {-5}{2}}}
Cet exercice est tombé au brevet des collèges (1996).
Les résultats seront donnés sous forme fractionnaire.
Que faut-il ajouter à 3 7 {\displaystyle {\frac {3}{7}}}
Solution
x + 3 7 = 2 {\displaystyle x+{\frac {3}{7}}=2}
x = 2 − 3 7 {\displaystyle x=2-{\frac {3}{7}}}
x = 14 7 − 3 7 {\displaystyle x={\frac {14}{7}}-{\frac {3}{7}}}
x = 14 − 3 7 {\displaystyle x={\frac {14-3}{7}}}
x = 11 7 {\displaystyle x={\frac {11}{7}}}
Que faut-il ajouter à 1 2 + 1 4 + 1 6 {\displaystyle {\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{6}}}
Solution
x + 1 2 + 1 4 + 1 6 = 1 {\displaystyle x+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{6}}=1}
x = 1 − 1 2 − 1 4 − 1 6 {\displaystyle x=1-{\frac {1}{2}}-{\frac {1}{4}}-{\frac {1}{6}}}
x = 12 12 − 6 12 − 3 12 − 2 12 {\displaystyle x={\frac {12}{12}}-{\frac {6}{12}}-{\frac {3}{12}}-{\frac {2}{12}}}
x = 12 − 6 − 3 − 2 12 {\displaystyle x={\frac {12-6-3-2}{12}}}
x = 1 12 {\displaystyle x={\frac {1}{12}}}
À un nombre, j'ajoute 7 5 {\displaystyle {\frac {7}{5}}} 3 11 {\displaystyle {\frac {3}{11}}}
Solution
( x + 7 5 ) × 3 11 = 1 {\displaystyle \left(x+{\frac {7}{5}}\right)\times {\frac {3}{11}}=1}
( x + 7 5 ) = 1 ÷ 3 11 {\displaystyle \left(x+{\frac {7}{5}}\right)=1\div {\frac {3}{11}}}
( x + 7 5 ) = 1 × 11 3 {\displaystyle \left(x+{\frac {7}{5}}\right)=1\times {\frac {11}{3}}}
( x + 7 5 ) = 3 3 × 11 3 {\displaystyle \left(x+{\frac {7}{5}}\right)={\frac {3}{3}}\times {\frac {11}{3}}}
( x + 7 5 ) = 3 × 11 3 × 3 {\displaystyle \left(x+{\frac {7}{5}}\right)={\frac {3\times 11}{3\times 3}}}
( x + 7 5 ) = 33 9 {\displaystyle \left(x+{\frac {7}{5}}\right)={\frac {33}{9}}}
x + 7 5 = 33 9 {\displaystyle x+{\frac {7}{5}}={\frac {33}{9}}}
x = 33 9 − 7 5 {\displaystyle x={\frac {33}{9}}-{\frac {7}{5}}}
x = 33 × 5 9 × 5 − 7 × 9 5 × 9 {\displaystyle x={\frac {33\times 5}{9\times 5}}-{\frac {7\times 9}{5\times 9}}}
x = 165 45 − 63 45 {\displaystyle x={\frac {165}{45}}-{\frac {63}{45}}}
x = 165 − 63 45 {\displaystyle x={\frac {165-63}{45}}}
x = 102 45 {\displaystyle x={\frac {102}{45}}}
x = 34 15 {\displaystyle x={\frac {34}{15}}}
Cet exercice est tombé au brevet des collèges (1996).
Calculer puis simplifier A = 13 14 − 1 15 × 10 7 {\displaystyle A={\frac {13}{14}}-{\frac {1}{15}}\times {\frac {10}{7}}}
Solution
A = 13 14 − 1 15 × 10 7 {\displaystyle A={\frac {13}{14}}-{\frac {1}{15}}\times {\frac {10}{7}}}
A = 13 14 − 1 × 10 15 × 7 {\displaystyle A={\frac {13}{14}}-{\frac {1\times 10}{15\times 7}}}
A = 13 14 − 10 105 {\displaystyle A={\frac {13}{14}}-{\frac {10}{105}}}
A = 13 14 − 2 21 {\displaystyle A={\frac {13}{14}}-{\frac {2}{21}}}
A = 13 × 21 14 × 21 − 2 × 14 21 × 14 {\displaystyle A={\frac {13\times 21}{14\times 21}}-{\frac {2\times 14}{21\times 14}}}
A = 273 294 − 28 294 {\displaystyle A={\frac {273}{294}}-{\frac {28}{294}}}
A = 273 − 28 294 {\displaystyle A={\frac {273-28}{294}}}
A = 245 294 {\displaystyle A={\frac {245}{294}}}
A = 35 42 {\displaystyle A={\frac {35}{42}}}
Cet exercice est tombé au brevet des collèges (1996).
Calculer et mettre le résultat sous la forme de fraction irréductible
A = 3 14 + 5 21 {\displaystyle A={\frac {3}{14}}+{\frac {5}{21}}}
Solution
A = 3 14 + 5 21 {\displaystyle A={\frac {3}{14}}+{\frac {5}{21}}}
A = 18 84 + 20 84 {\displaystyle A={\frac {18}{84}}+{\frac {20}{84}}}
A = 18 + 20 84 {\displaystyle A={\frac {18+20}{84}}}
A = 38 84 {\displaystyle A={\frac {38}{84}}}
A = 19 42 {\displaystyle A={\frac {19}{42}}}
B = 2 3 − 7 3 × 9 34 {\displaystyle B={\frac {2}{3}}-{\frac {7}{3}}\times {\frac {9}{34}}}
Solution
B = 2 3 − 7 3 × 9 34 {\displaystyle B={\frac {2}{3}}-{\frac {7}{3}}\times {\frac {9}{34}}}
B = 2 3 − 7 × 9 3 × 34 {\displaystyle B={\frac {2}{3}}-{\frac {7\times 9}{3\times 34}}}
B = 2 3 − 63 102 {\displaystyle B={\frac {2}{3}}-{\frac {63}{102}}}
B = 68 102 − 63 102 {\displaystyle B={\frac {68}{102}}-{\frac {63}{102}}}
B = 68 − 63 102 {\displaystyle B={\frac {68-63}{102}}}
B = 5 102 {\displaystyle B={\frac {5}{102}}}
C = 2 3 3 2 ÷ 2 4 3 {\displaystyle C={\frac {2^{3}}{3^{2}}}\div {\frac {2^{4}}{3}}}
Solution
C = 2 3 3 2 ÷ 2 4 3 {\displaystyle C={\frac {2^{3}}{3^{2}}}\div {\frac {2^{4}}{3}}}
C = 2 3 3 2 × 3 2 4 {\displaystyle C={\frac {2^{3}}{3^{2}}}\times {\frac {3}{2^{4}}}}
C = 2 3 2 4 × 3 3 2 {\displaystyle C={\frac {2^{3}}{2^{4}}}\times {\frac {3}{3^{2}}}}
C = 1 2 × 1 3 {\displaystyle C={\frac {1}{2}}\times {\frac {1}{3}}}
C = 1 2 × 3 {\displaystyle C={\frac {1}{2\times 3}}}
C = 1 6 {\displaystyle C={\frac {1}{6}}}
