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}}}
