corrigé

◦
10−4= 0,0001 10−4/∈]− ∞; 0[
5>x > 2x∈]2 ; 5] x∈[5 ; 2[ [5 ; 2[ 5 >2
N[0 ; +∞[
4,5∈[0 ; +∞[ 4,5/∈N
I= [−3 ; +∞[J=] − ∞; 0]
I∪J=] − ∞; +∞[= R
2< x > 5
x∈[2 ; +∞[∩]5 ; +∞[ 2 < x > 5x∈[2 ; +∞[ 2 6x
4>0>
4∈[0 ; +∞[
7>5 7 ∈[5 ; +∞[ ] −∞;−5] [5 ; +∞[
7∈]− ∞;−5] ∪[5 ; +∞[
I⊂J I ∩J I J
I J J I ∩J
I I = [0 ; +∞[J= [−5 ; +∞[I∩J= [0 ; +∞[
I∩J̸=J
5
3= 1,66666...... 065 3<1,66667 5 3∈[0 ; 1,66667[
[0 ; 1]
0,158
[0 ; 1]

A= 5 −2√9−4 + 5
6÷1 3−1 2+2√52= 5 −2√5 + 5 6÷2−3 6+ 4 ×5 = 5 −2√5 + 5 6÷−1 6+ 20
= 25 −2√5−5
6×6 = 25 −2√5−5 = 20 −2√5
1260 ×0,98 = 1234,8
1200 + 60 ×5 = 1500
1−0,02 ×5 = 0,9 1500 ×0,9 = 1350
1200 + 10 ×60 = 1800
1−10 ×0,02 = 0,8 1800 ×0,8 = 1440
n
1200 + 60n
1−0,02n
R(n) = (1200 + 60n)(1 −0,02n) = 1200 −1200 ×0,02n+ 60n−60n×0,02n
= 1200 −24n+ 60n−1,2n2=−1,2n2+ 36n+ 1200
n= 15
1,7×400
0,45 ≃1510
n= 15 R(15) R(15) = −1,2×152+ 36 ×15 + 1200 = 1470
n= 30
R(n) = 1200 ⇔ −1,2n2+ 36n+ 1200 = 1200 ⇔ −1,2n2+ 36n= 0 ⇔n(−1,2n+ 36) = 0
n= 0 −1,2n+ 36 = 0 −1,2n=−36
n=−36
−12
10
=36
6
5
= 36 ×5
6= 6 ×5 = 30
x
x
AB = 3 AC = 5
BC AC2=AB2+BC2BC2= 25 −9 = 16 BC =√16 = 4
0< x < 4x∈]0 ; 4[

A H A
B C
x BH 0< x < BH
H
[BC]BH =10
2= 5 0 < x < 5x∈]0 ; 5[
ABCD A B
x[AC]
0< x < AC AC ACD
D AC2=AD2+DC2= 102+ 102= 100 + 100 = 200 AC =√200 = √2×√100 = 10√2
0< x < 10√2x∈]0 ; 10√2[