Fraction/Exercices/Sujet de brevet

Exercice 1

Calculer et donner le résultat sous la forme d'une fraction la plus simple possible :

$A={\frac {4}{3}}-{\frac {1}{3}}\times \left(3+{\frac {1}{2}}\right)$

Solution

A = ${\frac {4}{3}}-{\frac {1}{3}}\times \left(3+{\frac {1}{2}}\right)$

= ${\frac {4}{3}}-{\frac {1}{3}}\times \left({\frac {6}{2}}+{\frac {1}{2}}\right)$

= ${\frac {4}{3}}-{\frac {1}{3}}\times \left({\frac {6+1}{2}}\right)$

= ${\frac {4}{3}}-{\frac {1}{3}}\times {\frac {7}{2}}$

= ${\frac {4}{3}}-{\frac {7}{6}}$

= ${\frac {8}{6}}-{\frac {7}{6}}$

= ${\frac {1}{6}}$

 

Exercice 2

Donner ces résultats sous formes de fraction irréductibles :

$A=\left({\frac {5}{7}}\right)^{2}-{\frac {2}{7}}$

Solution

$A=\left({\frac {5}{7}}\right)^{2}-{\frac {2}{7}}$

= $\left({\frac {25}{49}}\right)-{\frac {2}{7}}$

= $\left({\frac {25}{49}}\right)-{\frac {2\times 7}{7\times 7}}$

= ${\frac {25}{49}}-{\frac {14}{49}}$

= ${\frac {25-14}{49}}$

= ${\frac {11}{49}}$

 

$B={\frac {1}{9}}+{\frac {1}{12}}$

Solution

$B={\frac {1}{9}}+{\frac {1}{12}}$

$={\frac {1\times 12}{9\times 12}}+{\frac {1\times 9}{12\times 9}}$

$={\frac {12}{108}}+{\frac {9}{108}}$

$={\frac {12+9}{108}}$

$={\frac {21}{108}}$

 

En électricité, pour calculer des valeurs de résistances, on utilise la formule: ${\frac {1}{R}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}$

Sachant que $R_{1}$=9 ohms et $R_{2}$= 12 ohms, déterminer la valeur exacte de R.

Solution

${\frac {1}{R}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}$

= ${\frac {1}{R}}={\frac {1}{9}}+{\frac {1}{12}}$

= ${\frac {1}{R}}={\frac {21}{108}}$

$R={\frac {108}{21}}ohms$

 

Exercice 3

Mettre sous la forme de fraction irréductible ${\frac {{\frac {11}{3}}-7}{\frac {25}{6}}}$

Solution

${\frac {{\frac {11}{3}}-7}{\frac {25}{6}}}$

= ${\frac {{\frac {11\times 2}{3\times 2}}-{\frac {7\times 6}{6}}}{\frac {25}{6}}}$

= ${\frac {{\frac {22}{6}}-{\frac {42}{6}}}{\frac {25}{6}}}$

= ${\frac {22-42}{25}}$

= $-{\frac {20}{25}}$

= $-{\frac {4}{5}}$

 

Mettre sous la forme de fraction irréductible ${\frac {3}{7}}-{\frac {2}{5}}\times {\frac {15}{4}}$

Solution

${\frac {3}{7}}-{\frac {2}{5}}\times {\frac {15}{4}}$

= ${\frac {3}{7}}-{\frac {2\times 15}{5\times 4}}$

= ${\frac {3\times 20}{7\times 20}}-{\frac {2\times 15\times 7}{5\times 4\times 7}}$

= ${\frac {60}{140}}-{\frac {210}{140}}$

= ${\frac {60-210}{140}}$

= $-{\frac {150}{140}}$

= $-{\frac {15}{14}}$

 

Exercice 4

calculer $A=\left({\frac {1}{3}}-{\frac {1}{5}}\right)\div {\frac {2}{5}}$

Solution

$A=\left({\frac {1}{3}}-{\frac {1}{5}}\right)\div {\frac {2}{5}}$

=$\left({\frac {1\times 5}{3\times 5}}-{\frac {1\times 3}{5\times 3}}\right)\div {\frac {2}{5}}$

=$\left({\frac {5}{15}}-{\frac {3}{15}}\right)\div {\frac {2}{5}}$

=${\frac {5-3}{15}}\div {\frac {2}{5}}$

=${\frac {\frac {2}{15}}{\frac {2}{5}}}$

=${\frac {2}{15}}\times {\frac {5}{2}}$

=${\frac {2\times 5}{15\times 2}}$

=${\frac {10}{30}}$

=${\frac {1}{3}}$

 

Exercice 5

Calculer $\left((-4+3)\times {\frac {2}{7}}\right)\div {\frac {3}{14}}$

Solution

$\left((-4+3)\times {\frac {2}{7}}\right)\div {\frac {3}{14}}$

= $\left(-1\times {\frac {2}{7}}\right)\div {\frac {3}{14}}$

= $\left(-{\frac {2}{7}}\right)\div {\frac {3}{14}}$

= $-{\frac {2}{7}}\times {\frac {14}{3}}$

= $-{\frac {2\times 14}{7\times 3}}$

= $-{\frac {2\times 2\times 7}{7\times 3}}$

= $-{\frac {2\times 2}{3}}$

= $-{\frac {4}{3}}$

 

Calculer ${\frac {4-\left(2-5\right)^{2}}{4+5}}$

Solution

${\frac {4-\left(2-5\right)^{2}}{4+5}}$

= ${\frac {4-\left(-3\right)^{2}}{9}}$

= ${\frac {4-9}{9}}$

= ${\frac {-5}{9}}$

 

Exercice 6

Calculer $3-3\div {\frac {9}{2}}$

Solution

$3-3\div {\frac {9}{2}}$

= $3-3\times {\frac {2}{9}}$

= $3-{\frac {3\times 2}{9}}$

= $3-{\frac {2}{3}}$

= ${\frac {9}{3}}-{\frac {2}{3}}$

= ${\frac {9-2}{3}}$

= ${\frac {7}{3}}$

 

Exercice 7

Calculer $A={\frac {7}{18}}\times {\frac {2}{7}}-\left({\frac {5}{3}}-1\right)^{2}$

Solution

$A={\frac {7}{18}}\times {\frac {2}{7}}-\left({\frac {5}{3}}-1\right)^{2}$

= ${\frac {7\times 2}{18\times 7}}-\left({\frac {5}{3}}-{\frac {3}{3}}\right)^{2}$

= ${\frac {2}{18}}-\left({\frac {5-3}{3}}\right)^{2}$

= ${\frac {1}{9}}-\left({\frac {2}{3}}\right)^{2}$

= ${\frac {1}{9}}-\left({\frac {4}{9}}\right)$

= $-{\frac {3}{9}}$

= $-{\frac {1}{3}}$

 

Exercice 8

$A={\frac {5}{7}}-{\frac {2}{7}}\times {\frac {4}{3}}$

Solution

$A={\frac {5}{7}}-{\frac {2}{7}}\times {\frac {4}{3}}$

= ${\frac {5\times 3}{7\times 3}}-{\frac {2\times 4}{7\times 3}}$

= ${\frac {15}{21}}-{\frac {8}{21}}$

= ${\frac {15-8}{21}}$

= ${\frac {7}{21}}$

= ${\frac {1}{3}}$

 

$B=\left({\frac {1}{2}}-{\frac {1}{3}}\right)\times {\frac {3}{2}}+1$

Solution

$B=\left({\frac {1}{2}}-{\frac {1}{3}}\right)\times {\frac {3}{2}}+1$

= $\left({\frac {3}{6}}-{\frac {2}{6}}\right)\times {\frac {3}{2}}+1$

= $\left({\frac {1}{6}}\right)\times {\frac {3}{2}}+1$

= ${\frac {3}{12}}+1$

= ${\frac {15}{12}}$

= ${\frac {5}{4}}$

 

Exercice 9

$A={\frac {3}{2}}-{\frac {5}{2}}\times {\frac {3}{10}}$

Solution

$A={\frac {3}{2}}-{\frac {5}{2}}\times {\frac {3}{10}}$

= ${\frac {3}{2}}-{\frac {15}{20}}$

= ${\frac {30}{20}}-{\frac {15}{20}}$

= ${\frac {15}{20}}$

= ${\frac {3}{4}}$

 

$B=\left({\frac {3}{5}}\right)^{2}\div {\frac {9}{20}}$

Solution

$B=\left({\frac {3}{5}}\right)^{2}\div {\frac {9}{20}}$

= $\left({\frac {9}{25}}\right)\times {\frac {20}{9}}$

= ${\frac {20}{25}}$

= ${\frac {4}{5}}$

 

Exercice 10

On pose $A=4-{\frac {3}{4}}\left({\frac {1}{3}}-{\frac {1}{6}}\right)$

En faisant apparaître les étapes de calcul, donner une écriture fractionnaire et une écriture décimale du nombre A.

Solution

$A=4-{\frac {3}{4}}\left({\frac {1}{3}}-{\frac {1}{6}}\right)$

$=4-{\frac {3}{4}}\left({\frac {2}{6}}-{\frac {1}{6}}\right)$

$=4-{\frac {3}{4}}\left({\frac {1}{6}}\right)$

$=4-{\frac {3}{24}}$

$=4-{\frac {1}{8}}$

$={\frac {32}{8}}-{\frac {1}{8}}$

$={\frac {31}{8}}$

$=3,875\,$

On retire une petite partie de 4, donc on sait que le résultat est relativement proche de 4 au final.

 

Exercice 11

Donner la valeur exacte la plus simple possible en indiquant le détail des calculs :

${\frac {34}{5}}\div \left({\frac {4}{5}}-{\frac {3}{8}}\right)$

Solution

$={\frac {34}{5}}\div \left({\frac {4}{5}}\times {\frac {8}{8}}-{\frac {3}{8}}\times {\frac {5}{5}}\right)$

$={\frac {34}{5}}\div \left({\frac {4}{5}}\times {\frac {8}{8}}-{\frac {3}{8}}\times {\frac {5}{5}}\right)$

$={\frac {34}{5}}\div \left({\frac {32-15}{5\times 8}}\right)$

$={\frac {34}{5}}\div \left({\frac {17}{5\times 8}}\right)$

$={\frac {34}{5}}\times \left({\frac {5\times 8}{17}}\right)$

$={\frac {17\times 2}{5}}\times \left({\frac {5\times 8}{17}}\right)$

$=2\times 8=16$

 

${\frac {24\times 10^{-2}\times 3,5\times 10^{5}}{8\times 10^{-1}\times 21\times 10^{4}}}$

Solution

$={\frac {24\times 3,5}{8\times 21}}\times {\frac {10^{-2}\times 10^{5}}{10^{-1}\times 10^{4}}}$

$={\frac {8\times 3\times 3,5}{8\times 3\times 7}}\times {\frac {10^{3}}{10^{3}}}$

$={\frac {3,5}{7}}$

$={\frac {1}{2}}$

 

Exercice 12

Annale de sujet d'examen

Cet exercice est tombé au brevet des collège (1995).

On considère les nombres :

• $A={\frac {6}{7}}-{\frac {4}{7}}\times {\frac {5}{2}}$

• $B={\frac {{\frac {3}{4}}-4}{{\frac {3}{4}}+{\frac {1}{3}}}}$

• $C=3^{2}\times 2-125\times 10^{-1}$

En précisant les différentes étapes des calculs

a) Écrire A sous la forme la plus simple possible et sans utiliser de valeur approchée

Solution

$A={\frac {6}{7}}-{\frac {4}{7}}\times {\frac {5}{2}}$

$A={\frac {6}{7}}-{\frac {4\times 5}{7\times 2}}$

$A={\frac {6}{7}}-{\frac {20}{14}}$

$A={\frac {12}{14}}-{\frac {20}{14}}$

$A={\frac {12-20}{14}}$

$A={\frac {-8}{14}}$

$A={\frac {-4}{7}}$

 

b) Écrire B sous la forme d'une nombre entier relatif

Solution

• $B={\frac {{\frac {3}{4}}-4}{{\frac {3}{4}}+{\frac {1}{3}}}}$

• $B={\frac {{\frac {3}{4}}-{\frac {16}{4}}}{{\frac {9}{12}}+{\frac {4}{12}}}}$

• $B={\frac {\frac {3-16}{4}}{\frac {9+4}{12}}}$

• $B={\frac {\frac {-13}{4}}{\frac {13}{12}}}$

• $B={\frac {-13}{4}}\times {\frac {12}{13}}$

• $B=-{\frac {12}{4}}$

• $B=-3$

 

c) Écrire C sous la forme d'un nombre décimal

Solution

• $C=3^{2}\times 2-125\times 10^{-1}$

• $C=9\times 2-12,5$

• $C=18-12,5$

• $C=5,5$

 

Exercice 13

Annale de sujet d'examen

Cet exercice est tombé au brevet des collège (1995).

Sachant que $a={\frac {2}{3}}$, $b=-{\frac {1}{4}}$, $c={\frac {2}{5}}$ et $d=-{\frac {1}{2}}$,

a) Calculer $A=ab+cd$

Solution

$A=ab+cd$

$A={\frac {2}{3}}\times {\frac {-1}{4}}+{\frac {2}{5}}\times {\frac {-1}{2}}$

$A={\frac {-2}{12}}+{\frac {-2}{10}}$

$A={\frac {-10}{60}}+{\frac {-12}{60}}$

$A={\frac {-22}{60}}$

$A={\frac {-11}{30}}$

 

b) Calculer $B={\frac {a+d}{b+c}}$

Solution

$B={\frac {a+d}{b+c}}$

$B={\frac {{\frac {2}{3}}-{\frac {1}{2}}}{-{\frac {1}{4}}+{\frac {2}{5}}}}$

$B={\frac {{\frac {4}{6}}-{\frac {3}{6}}}{{\frac {-5}{20}}+{\frac {8}{20}}}}$

$B={\frac {\frac {1}{6}}{\frac {3}{20}}}$

$B={\frac {1\times 20}{6\times 3}}$

$B={\frac {20}{18}}$

$B={\frac {10}{9}}$

 

Exercice 14

Annale de sujet d'examen

Cet exercice est tombé au brevet des collège (1995).

Calculer et donner chaque résultat sous la forme d'une fraction aussi simple que possible

a) $A={\frac {3}{4}}-{\frac {5}{6}}\times {\frac {3}{2}}$

Solution

$A={\frac {3}{4}}-{\frac {5}{6}}\times {\frac {3}{2}}$

$A={\frac {3}{4}}-{\frac {5\times 3}{6\times 2}}$

$A={\frac {3}{4}}-{\frac {15}{12}}$

$A={\frac {9}{12}}-{\frac {15}{12}}$

$A={\frac {9-15}{12}}$

$A={\frac {-6}{12}}$

$A=-{\frac {1}{2}}$