Entraînement

Maîtriser les fractions

Signaler

Énoncé

Exercice 1

Ecrire chacun des nombres suivants sous la forme d'une fraction irréductible :

D=587+5D = \dfrac{5 - 8}{7 + 5}
E=(3412×25)×(27)2E = \left(\dfrac{3}{4} - \dfrac{1}{2} \times \dfrac{2}{5}\right) \times \left(\dfrac{2}{7}\right)^2
F=(23115)249F = \dfrac{\left(\dfrac{2}{3} - \dfrac{1}{15}\right)^2}{\dfrac{4}{9}}

Exercice 2

M=314221M = \dfrac{3}{14} - \dfrac{2}{21}
N=518+127N = \dfrac{5}{18} + \dfrac{1}{27}
O=124130O = \dfrac{1}{24} - \dfrac{-1}{30}
P=19+115P = \dfrac{-1}{9} + \dfrac{1}{-15}

Q=1942+22113Q = \dfrac{19}{42} + \dfrac{2}{21} - \dfrac{1}{3}
R=1742+22113R = \dfrac{17}{42} + \dfrac{2}{21} - \dfrac{1}{3}
S=835235S = \dfrac{8}{35} - \dfrac{-2}{35}
T=38218T = \dfrac{3}{8} - 2 - \dfrac{-1}{8}

U=73×2714U = \dfrac{7}{3} \times \dfrac{27}{14}
V=85×1516V = \dfrac{-8}{5} \times \dfrac{-15}{16}
W=34:940W = \dfrac{3}{-4} : \dfrac{-9}{40}
X=169×314×58X = \dfrac{16}{9} \times \dfrac{3}{14} \times \dfrac{5}{8}

Exercice 3

Calculer les expressions et écrire les résultats sous la forme de fractions irréductibles.

M=(34)2M = \left(\dfrac{3}{4} \right)^2
N=324N = \dfrac{3^2}{4}
O=23+310×359O = \dfrac{2}{3} + \dfrac{3}{10} \times \dfrac{35}{9}
P=(23+310)×359P = \left(\dfrac{2}{3} + \dfrac{3}{10} \right) \times \dfrac{35}{9}

Q=(112):(214)Q = \left(1 - \dfrac{1}{2} \right) : \left(2 - \dfrac{1}{4} \right)
R=112:214R = 1 - \dfrac{1}{2} : 2 - \dfrac{1}{4}
S=34527S = \dfrac{\dfrac{3}{4} - \dfrac{5}{2}}{7}
T=(1913)2×92T = \left(\dfrac{1}{9} - \dfrac{1}{3}\right)^2 \times \dfrac{9}{2}

U=53791412U = \dfrac{\dfrac{5}{3} - \dfrac{7}{9}}{\dfrac{1}{4} - \dfrac{1}{2}}
V=152(15)2V = \dfrac{1 - 5^2}{(1 - 5)^2}
W=(67)2W = \left(\dfrac{6}{-7}\right)^2
X=523X = \dfrac{5^2}{-3}

Y=(110715):(76+716)Y = \left(\dfrac{1}{10} - \dfrac{7}{15}\right) : \left(\dfrac{7}{6} + \dfrac{7}{16}\right)
Z=18712:(76+716)Z = \dfrac{1}{8} - \dfrac{7}{12} : \left(\dfrac{7}{6} + \dfrac{7}{16}\right)
A=18+71256415A' = \dfrac{\dfrac{1}{8} + \dfrac{7}{12}}{\dfrac{5}{6} - \dfrac{4}{15}}

B=(18+712)×(65154)B' = \left(\dfrac{1}{8} + \dfrac{7}{12}\right) \times \left(\dfrac{6}{5} - \dfrac{15}{4}\right)

Exercice 4

  1. Écrire chacun des nombres suivants sous la forme d'une fraction irréductible :

A=15+3B=16+12C=11213D=113115A = \dfrac{1}{5+3} \hspace{25pt} B = \dfrac{1}{6} + \dfrac{1}{2} \hspace{25pt} C = \dfrac{1}{\dfrac{1}{2} - \dfrac{1}{3}} \hspace{25pt} D = \dfrac{1}{\dfrac{1}{3}} - \dfrac{1}{\dfrac{1}{5}}

  1. a) La somme des inverses de deux nombres est-elle égale à l'inverse de la somme de ces deux nombres ? (Justifier)
    b) La différence des inverses de deux nombres est-elle égale à l'inverse de la différence de ces deux nombres ? (Justifier)

Révéler le corrigé

Exercice 1

D=587+5=312D = \dfrac{5 - 8}{7 + 5} = \dfrac{-3}{12}

D=33×4=14D = -\dfrac{3}{3 \times 4} = -\dfrac{1}{4}

E=(3412×25)×(27)2=(341×22×5)×2272E = \left(\dfrac{3}{4} - \dfrac{1}{2} \times \dfrac{2}{5}\right) \times \left(\dfrac{2}{7}\right)^2 = \left(\dfrac{3}{4} - \dfrac{1 \times 2}{2 \times 5}\right) \times \dfrac{2^2}{7^2}

E=(3415)×449=(3×54×51×45×4)×449E = \left(\dfrac{3}{4} - \dfrac{1}{5}\right) \times \dfrac{4}{49} = \left(\dfrac{3 \times 5}{4 \times 5} - \dfrac{1 \times 4}{5 \times 4}\right) \times \dfrac{4}{49}

E=(1520420)×449=1120×449E = \left(\dfrac{15}{20} - \dfrac{4}{20}\right) \times \dfrac{4}{49} = \dfrac{11}{20} \times \dfrac{4}{49}

E=11×420×49=11×45×4×49E = \dfrac{11 \times 4}{20 \times 49} = \dfrac{11 \times 4}{5 \times 4 \times 49}

E=115×49=11245E = \dfrac{11}{5 \times 49} = \dfrac{11}{245}

F=(23115)249=(2×53×5115)249F = \dfrac{\left(\dfrac{2}{3} - \dfrac{1}{15}\right)^2}{\dfrac{4}{9}} = \dfrac{\left(\dfrac{2 \times 5}{3 \times 5} - \dfrac{1}{15}\right)^2}{\dfrac{4}{9}}

F=(1015115)249=(915)249F = \dfrac{\left(\dfrac{10}{15} - \dfrac{1}{15}\right)^2}{\dfrac{4}{9}} = \dfrac{\left(\dfrac{9}{15}\right)^2}{\dfrac{4}{9}}

F=92152×94F = \dfrac{9^2}{15^2} \times \dfrac{9}{4}

F=9×9×3×35×3×5×3×4F = \dfrac{9 \times 9 \times 3 \times 3}{5 \times 3 \times 5 \times 3 \times 4}

Exercice 2

M=314221=3×314×32×221×2M = \dfrac{3}{14} - \dfrac{2}{21} = \dfrac{3 \times 3}{14 \times 3} - \dfrac{2 \times 2}{21 \times 2}
M=942442=542M = \dfrac{9}{42} - \dfrac{4}{42} = \dfrac{5}{42}

N=518+127=5×318×3+1×227×2N = \dfrac{5}{18} + \dfrac{1}{27} = \dfrac{5 \times 3}{18 \times 3} + \dfrac{1 \times 2}{27 \times 2}
N=1554+254=1754N = \dfrac{15}{54} + \dfrac{2}{54} = \dfrac{17}{54}

O=124130=1×524×51×430×4O = \dfrac{1}{24} - \dfrac{-1}{30} = \dfrac{1 \times 5}{24 \times 5} - \dfrac{-1 \times 4}{30 \times 4}
O=5120+4120=9120O = \dfrac{5}{120} + \dfrac{4}{120} = \dfrac{9}{120}
O=340O = \dfrac{3}{40}

P=19+115=1×59×5+1×315×3P = \dfrac{-1}{9} + \dfrac{1}{-15} = \dfrac{-1 \times 5}{9 \times 5} + \dfrac{1 \times 3}{-15 \times 3}
P=545345=845P = -\dfrac{5}{45} - \dfrac{3}{45} = -\dfrac{8}{45}

Q=1942+22113=1942+2×221×21×143×14Q = \dfrac{19}{42} + \dfrac{2}{21} - \dfrac{1}{3} = \dfrac{19}{42} + \dfrac{2 \times 2}{21 \times 2} - \dfrac{1 \times 14}{3 \times 14}
Q=1942+4421442=942Q = \dfrac{19}{42} + \dfrac{4}{42} - \dfrac{14}{42} = \dfrac{9}{42}
Q=314Q = \dfrac{3}{14}

R=1742+22113=1742+4421442R = \dfrac{17}{42} + \dfrac{2}{21} - \dfrac{1}{3} = \dfrac{17}{42} + \dfrac{4}{42} - \dfrac{14}{42}
R=742=16R = \dfrac{7}{42} = \dfrac{1}{6}

S=835235=835+235S = \dfrac{8}{35} - \dfrac{-2}{35} = \dfrac{8}{35} + \dfrac{2}{35}
S=1035=27S = \dfrac{10}{35} = \dfrac{2}{7}

T=38218=382×88+18T = \dfrac{3}{8} - 2 - \dfrac{-1}{8} = \dfrac{3}{8} - \dfrac{2 \times 8}{8} + \dfrac{1}{8}
T=38168+18=128T = \dfrac{3}{8} - \dfrac{16}{8} + \dfrac{1}{8} = -\dfrac{12}{8}
T=4×34×2=32T = -\dfrac{4 \times 3}{4 \times 2} = -\dfrac{3}{2}

U=73×2714=7×9×33×7×2=92U = \dfrac{7}{3} \times \dfrac{27}{14} = \dfrac{7 \times 9 \times 3}{3 \times 7 \times 2} = \dfrac{9}{2}

V=85×1516=8×(15)5×16V = \dfrac{-8}{5} \times \dfrac{-15}{16} = \dfrac{-8 \times (-15)}{5 \times 16}
V=2×4×5×35×4×2×2=32V = \dfrac{2 \times 4 \times 5 \times 3}{5 \times 4 \times 2 \times 2} = \dfrac{3}{2}

W=34:940=34×409W = \dfrac{3}{-4} : \dfrac{-9}{40} = \dfrac{3}{-4} \times \dfrac{40}{-9}
W=3×4×104×3×3=103W = \dfrac{3 \times 4 \times 10}{4 \times 3 \times 3} = \dfrac{10}{3}

X=169×314×58=16×3×59×14×8X = \dfrac{16}{9} \times \dfrac{3}{14} \times \dfrac{5}{8} = \dfrac{16 \times 3 \times 5}{9 \times 14 \times 8}
X=2×2×4×3×53×3×7×2×4×2=53×7X = \dfrac{2 \times 2 \times 4 \times 3 \times 5}{3 \times 3 \times 7 \times 2 \times 4 \times 2} = \dfrac{5}{3 \times 7}
X=521X = \dfrac{5}{21}

Exercice 3

M=(34)2=3242M = \left(\dfrac{3}{4} \right)^2 = \dfrac{3^2}{4^2}
M=916M = \dfrac{9}{16}

N=324=94N = \dfrac{3^2}{4} = \dfrac{9}{4}

O=23+310×359=23+3×3510×3×3O = \dfrac{2}{3} + \dfrac{3}{10} \times \dfrac{35}{9} = \dfrac{2}{3} + \dfrac{3 \times 35}{10 \times 3 \times 3}
O=23+3530=2030+3530O = \dfrac{2}{3} + \dfrac{35}{30} = \dfrac{20}{30} + \dfrac{35}{30}
O=5530=5×115×6O = \dfrac{55}{30} = \dfrac{5 \times 11}{5 \times 6}
O=116O = \dfrac{11}{6}

P=(23+310)×359=(2030+930)×359P = \left(\dfrac{2}{3} + \dfrac{3}{10} \right) \times \dfrac{35}{9} = \left(\dfrac{20}{30} + \dfrac{9}{30} \right) \times \dfrac{35}{9}
P=2930×359=29×7×56×5×9P = \dfrac{29}{30} \times \dfrac{35}{9} = \dfrac{29 \times 7 \times 5}{6 \times 5 \times 9}
P=20354P = \dfrac{203}{54}

Q=(112):(214)=(2212):(8414)Q = \left(1 - \dfrac{1}{2} \right) : \left(2 - \dfrac{1}{4} \right) = \left(\dfrac{2}{2} - \dfrac{1}{2} \right) : \left(\dfrac{8}{4} - \dfrac{1}{4} \right)
Q=12:74=12×47Q = \dfrac{1}{2} : \dfrac{7}{4} = \dfrac{1}{2} \times \dfrac{4}{7}
Q=1×42×7=27Q = \dfrac{1 \times 4}{2 \times 7} = \dfrac{2}{7}

R=112:214=112×1214R = 1 - \dfrac{1}{2} : 2 - \dfrac{1}{4} = 1 - \dfrac{1}{2} \times \dfrac{1}{2} - \dfrac{1}{4}
R=11414=441414R = 1 - \dfrac{1}{4} - \dfrac{1}{4} = \dfrac{4}{4} - \dfrac{1}{4} - \dfrac{1}{4}
R=24=22×2R = \dfrac{2}{4} = \dfrac{2}{2 \times 2}
R=12R = \dfrac{1}{2}

S=34527=341047S = \dfrac{\dfrac{3}{4} - \dfrac{5}{2}}{7} = \dfrac{\dfrac{3}{4} - \dfrac{10}{4}}{7}
S=747=74×17S = \dfrac{-\dfrac{7}{4}}{7} = -\dfrac{7}{4} \times \dfrac{1}{7}
S=7×14×7=14S = -\dfrac{7 \times 1}{4 \times 7} = -\dfrac{1}{4}

T=(1913)2×92=(1939)2×92T = \left(\dfrac{1}{9} - \dfrac{1}{3}\right)^2 \times \dfrac{9}{2} = \left(\dfrac{1}{9} - \dfrac{3}{9}\right)^2 \times \dfrac{9}{2}
T=(29)2×92=481×92T = \left(-\dfrac{2}{9}\right)^2 \times \dfrac{9}{2} = \dfrac{4}{81} \times \dfrac{9}{2}
T=2×2×99×9×2=29T = \dfrac{2 \times 2 \times 9}{9 \times 9 \times 2} = \dfrac{2}{9}

U=53791412=159791424U = \dfrac{\dfrac{5}{3} - \dfrac{7}{9}}{\dfrac{1}{4} - \dfrac{1}{2}} = \dfrac{\dfrac{15}{9} - \dfrac{7}{9}}{\dfrac{1}{4} - \dfrac{2}{4}}
U=8914=89×4U = \dfrac{\dfrac{8}{9}}{-\dfrac{1}{4}} = -\dfrac{8}{9} \times 4
U=329U = -\dfrac{32}{9}

V=152(15)2=125(4)2V = \dfrac{1 - 5^2}{(1 - 5)^2} = \dfrac{1 - 25}{(-4)^2}
V=2416=3×82×8V = \dfrac{-24}{16} = \dfrac{-3 \times 8}{2 \times 8}
V=32V = \dfrac{-3}{2}

W=(67)2=62(7)2W = \left(\dfrac{6}{-7}\right)^2 = \dfrac{6^2}{(-7)^2}
W=3649W = \dfrac{36}{49}

X=523=253X = \dfrac{5^2}{-3} = -\dfrac{25}{3}

Y=(110715):(76+716)=(3301430):(5648+2148)Y = \left(\dfrac{1}{10} - \dfrac{7}{15}\right) : \left(\dfrac{7}{6} + \dfrac{7}{16}\right) = \left(\dfrac{3}{30} - \dfrac{14}{30}\right) : \left(\dfrac{56}{48} + \dfrac{21}{48}\right)
Y=1130:7748=1130×4877Y = -\dfrac{11}{30} : \dfrac{77}{48} = -\dfrac{11}{30} \times \dfrac{48}{77}
Y=11×2×3×82×3×5×7×11=85×7Y = -\dfrac{11 \times 2 \times 3 \times 8}{2 \times 3 \times 5 \times 7 \times 11} = -\dfrac{8}{5 \times 7}
Y=835Y = -\dfrac{8}{35}

Z=18712:(76+716)=18712:(5648+2148)Z = \dfrac{1}{8} - \dfrac{7}{12} : \left(\dfrac{7}{6} + \dfrac{7}{16}\right) = \dfrac{1}{8} - \dfrac{7}{12} : \left(\dfrac{56}{48} + \dfrac{21}{48}\right)
Z=18712:7748=18712×4877Z = \dfrac{1}{8} - \dfrac{7}{12} : \dfrac{77}{48} = \dfrac{1}{8} - \dfrac{7}{12} \times \dfrac{48}{77}
Z=187×12×412×7×11=18411Z = \dfrac{1}{8} - \dfrac{7 \times 12 \times 4}{12 \times 7 \times 11} = \dfrac{1}{8} - \dfrac{4}{11}
Z=11883288=2188Z = \dfrac{11}{88} - \dfrac{32}{88} = -\dfrac{21}{88}

A=18+71256415=324+14242530830A' = \dfrac{\dfrac{1}{8} + \dfrac{7}{12}}{\dfrac{5}{6} - \dfrac{4}{15}} = \dfrac{\dfrac{3}{24} + \dfrac{14}{24}}{\dfrac{25}{30} - \dfrac{8}{30}}
A=17241730=1724×3017A' = \dfrac{\dfrac{17}{24}}{\dfrac{17}{30}} = \dfrac{17}{24} \times \dfrac{30}{17}
A=17×2×3×52×4×3×17=54A' = \dfrac{17 \times 2 \times 3 \times 5}{2 \times 4 \times 3 \times 17} = \dfrac{5}{4}

B=(18+712)×(65154)=(324+1424)×(24207520)B' = \left(\dfrac{1}{8} + \dfrac{7}{12}\right) \times \left(\dfrac{6}{5} - \dfrac{15}{4}\right) = \left(\dfrac{3}{24} + \dfrac{14}{24}\right) \times \left(\dfrac{24}{20} - \dfrac{75}{20}\right)
B=1724×(5120)=17×5124×20B' = \dfrac{17}{24} \times \left(-\dfrac{51}{20}\right) = -\dfrac{17 \times 51}{24 \times 20}

B=17×3×173×8×20=289160B' = -\dfrac{17 \times 3 \times 17}{3 \times 8 \times 20} = -\dfrac{289}{160}

Exercice 4

1.1.

A=15+3=18 \text{A} = \dfrac{1}{5+3} = \dfrac{1}{8}

B=16+12=16+36=46=23 \text{B} = \dfrac{1}{6} + \dfrac{1}{2} = \dfrac{1}{6} + \dfrac{3}{6} = \dfrac{4}{6} = \dfrac{2}{3}

C=11213=13626=116=6 \text{C} = \dfrac{1}{\dfrac{1}{2} - \dfrac{1}{3}} = \dfrac{1}{\dfrac{3}{6} - \dfrac{2}{6}} = \dfrac{1}{\dfrac{1}{6}} = 6

D=113115=35=2 \text{D} = \dfrac{1}{\dfrac{1}{3}}-\dfrac{1}{\dfrac{1}{5}} = 3 - 5 = -2

2.2. a) Prenons deux nombres : 2 et 3.

La somme des inverses de ces deux nombres est :
12+13=36+26=56\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}

L'inverse de la somme de ces deux nombres est :
12+3=15\dfrac{1}{2 + 3} = \dfrac{1}{5}

Et 5615\dfrac{5}{6} \neq \dfrac{1}{5}, donc 12+1312+3\dfrac{1}{2} + \dfrac{1}{3} \neq \dfrac{1}{2 + 3}.

Conclusion : La somme des inverses de deux nombres n'est pas égale à l'inverse de la somme de ces deux nombres.

2.2. b) Prenons deux nombres : 2 et 3.

La différence des inverses de ces deux nombres est :
1213=3626=16\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{3}{6} - \dfrac{2}{6} = \dfrac{1}{6}

L'inverse de la différence de ces deux nombres est :
123=1\dfrac{1}{2-3} = -1

Et 161\dfrac{1}{6} \neq -1, donc 1213123\dfrac{1}{2} - \dfrac{1}{3} \neq \dfrac{1}{2-3}.

Conclusion : La différence des inverses de deux nombres n'est pas égale à l'inverse de la différence de ces deux nombres.