تمارين - TCSF & TCTF

الجذع المشترك العلمي و التكنولوجي – خيار فرنسية


درس : Ensemble de nombres et calcul dans R

Exercice 1

  1. Calculer : A=53+16143A=\dfrac53+\dfrac16-\dfrac14-3 et B=311+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+bc)2(ab2c)+6(2ab)E=3(a+b-c)-2(a-b-2c)+6(2a-b)
F=3a22[(a2b)3(3a+2b)]F=3a-2-2[(a-2b)-3(3a+2b)]
  • Pour A
A=53+16143=4×5+1×21×33×1212=20+233612=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

311+12=312+12=323=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+bc)2(ab2c)+6(2ab)=3a+3b3c2a+2b+4c+12a6b=(32+12)a+(3+26)b+(3+4)c=13ab+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=3a22[(a2b)3(3a+2b)]=3a22[a2b9a6b]=3a22[8a8b]=3a2+16a+16b=19a+16b2\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*}