📝 Exercices de la série▼
Exercice 1Exercice 2Exercice 3Exercice 4Exercice 5Exercice 6Exercice 7Exercice 8Exercice 9Exercice 10Exercice 11Exercice 12Exercice 13

Exercice 1

  1. Calculer : A=53+16−14−3A=\dfrac53+\dfrac16-\dfrac14-3 et B=3−11+121+23B=\dfrac{3-\dfrac{1}{1+\dfrac{1}{2}}}{1+\dfrac{2}{3}}
  2. Soient a, b et c des réels, simplifier:
E=3(a+b−c)−2(a−b−2c)+6(2a−b)E=3(a+b-c)-2(a-b-2c)+6(2a-b) F=3a−2−2[(a−2b)−3(3a+2b)]F=3a-2-2[(a-2b)-3(3a+2b)]
  • Pour A
A=53+16−14−3=4×5+1×2−1×3−3×1212=20+2−3−3612=−1712\begin{align*} A&=\dfrac53+\dfrac16-\dfrac14-3\\ &=\dfrac{4\times5+1\times2-1\times3-3\times12}{12}\\ &=\dfrac{20+2-3-36}{12} \\ &=-\dfrac{17}{12} \end{align*}
  • Pour B

on a

3−11+12=3−12+12=3−23=78\begin{align*} 3-\dfrac{1}{1+\dfrac{1}{2}} = 3-\dfrac{1}{\dfrac{2+1}{2}} =3-\dfrac{2}{3} =\dfrac{7}{8} \end{align*} 1+23=1+3+23=1+53=381+\dfrac{2}{3} = 1+\dfrac{3+2}{3} = 1+\dfrac{5}{3} = \dfrac{3}{8} B=7383=73×38=78\begin{align*} B=\dfrac{\dfrac{7}{3}}{\dfrac{8}{3}} =\dfrac{7}{3}\times\dfrac{3}{8} =\dfrac{7}{8} \end{align*}
E=3(a+b−c)−2(a−b−2c)+6(2a−b)=3a+3b−3c−2a+2b+4c+12a−6b=(3−2+12)a+(3+2−6)b+(−3+4)c=13a−b+c\begin{align*} E&=3(a+b-c)-2(a-b-2c)+6(2a-b) \\ &=3a+3b-3c-2a+2b+4c+12a-6b \\ &=(3-2+12)a+(3+2-6)b+(-3+4)c \\ &=13a-b+c \end{align*} F=3a−2−2[(a−2b)−3(3a+2b)]=3a−2−2[a−2b−9a−6b]=3a−2−2[−8a−8b]=3a−2+16a+16b=19a+16b−2\begin{align*} F&=3a-2-2[(a-2b)-3(3a+2b)] \\ &=3a-2-2[a-2b-9a-6b] \\ &=3a-2-2[-8a-8b] \\ &=3a-2+16a+16b \\ &=19a+16b-2 \end{align*}