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Description
Sujets
Informations
Publié par | bankexam |
Publié le | 27 janvier 2008 |
Nombre de lectures | 26 |
Langue | Français |
Extrait
16 2007 2 4 (un
) +∞ (un
) +∞ (un
) +∞ C C z
2
+(3+i)z+
7
4
+i=0.
H ={(x,y)∈R
2
,x
2
+y
2
=1} R
2
(R
2
,+) x y∈Q xy∈Q x∈R−Q y∈Q xy∈R−Q xy∈Q x∈R−Q y∈R−Q xy∈R−Q 7 ab=ba (a+b)
n
=
P
n
k=0
C
k
n
a
k
b
n−k
C
k
n
=
n!
k!(n−k)!
A =
0
@
1 1 1
0 1 1
0 0 1
1
A
n∈N A
n
A=
0
@
1 0 0
0 1 0
0 0 1
1
A
+
0
@
0 1 1
0 0 1
0 0 0
1
A
B =
0
@
2 0 1
0 3 0
0 0 2
1
A
B
n
n∈N 7 F F
11 2007 16 2007 2 C C z
1 =a
1 +ib
1
z
2 =a
2 +ib
2
z
1 „z
2
a
1 <a
2
(a
1 =a
2
b
16b
2
).
2+5i„ 3+i 2 < 3 2+i„ 2+5i 2 = 2 36 5 3+7i6„2+8i „ C „ z
1
,z
2
,z
3 ∈C z
1 „z
2 ⇒z
1 +z
3 „z
2 +z
3
.
0„i „ (C,„) „ 7 11 2007 16 2007 2 (un
) un = −
1
n
n ∈ N
∗
limn→+∞un =0 (un
) u
2k = 2k u
2k+1 = 2k−1 limn→+∞un =+∞ +∞ (un
) M > 0 N ∈N uN >M n>N un>uN >M M >0 N ∈N n>N ⇒un >M limn→∞un =+∞ z
1 =
−3+
q
√
5+1
2
+i(−1+
q
√
5−1
2
)
2
z
1 =
−3−
q
√
5+1
2
+i(−1−
q
√
5−1
2
)
2
H R
2
(0,0) R
2
H x,y ∈ Q p
1
,p
2 ∈ Z q
1
,q
2 ∈ N
∗
x =
p1
q1
y =
p2
q2
xy =
p1p2
q1q2
∈Q 1 y 6= 0 x ∈ R−Q y ∈ Q
∗
xy ∈ R−Q y∈Q
∗
xy∈Q p∈Z
∗
r∈Z q,s∈N y =
p
q
xy =
r
s
p6=0 x=
q
p
×
r
s
∈Q 2 y =0 y =0 xy =0∈Q x=
√
2∈R−Q y =
√
2∈R−Q xy =2∈Q A = C +D C =
0
@
1 0 0
0 1 0
0 0 1
1
A
D =
0
@
0 1 1
0 0 1
0 0 0
1
A
CD = DC
C =I
3
A
n
=(C+D)
n
=C
n
+nC
n−1
D+
n(n−1)
2
C
n−2
D
2
+
n(n−1)(n−2)
6
C
n−3
D
3
+···+D
n
.
D
2
=
0
@
0 0 1
0 0 0
0 0 0
1
A
D
n
=(0) n>3 C
k
=C A
n
=
0
@
1 0 0
0 1 0
0 0 1
1
A
+n
0
@
0 1 1
0 0 1
0 0 0
1
A
+
n(n−1)
2
0
@
0 0 1
0 0 0
0 0 0
1
A
=
0
@
1 n
n(n+1)
2
0 1 n
0 0 1
1
A
11 2007 16 2007 2 B =
0
@
2 0 1
0 3 0
0 0 2
1
A
B =E+F E =
0
@
2 0 0
0 3 0
0 0 2
1
A
F =
0
@
0 0 1
0 0 0
0 0 0
1
A
EF =FE E
n
=
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
n=0 E
0
= I
3
n E
n
=
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
E
n+1
= E
n
×E =
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
×
0
@
2 0 0
0 3 0
0 0 2
1
A
=
0
@
2
n
×2 0 0
0 3
n
×3 0
0 0 2
n
×2
1
A
=
0
@
2
n+1
0 0
0 3
n+1
0
0 0 2
n+1
1
A
n+1 n∈N F
n
=(0) n>2 B
n
=E
n
+nE
n−1
F
=
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
+n
0
@
2
n−1
0 0
0 3
n−1
0
0 0 2
n−1
1
A
0
@
0 0 1
0 0 0
0 0 0
1
A
=
0
@
2
n
0 n2
n−1
0 3
n
0
0 0 2
n
1
A
„ z =a+ib∈C a=a b6b z„z z
1 = a
1 +ib
1
z
2 = a
2 +ib
2
z
1 „ z
2
z
2 „ z
1
a
1 < a
2
a
1 = a
2
b
1 6 b
2
a
1 < a
2
z
2 „ z
1
a
1 = a
2
‰
z
1 „z
2 ⇒ b
16b
2
z
2 „z
1 ⇒ b
26b
1
b
1 =b
2
z
1 =z
2
z
1 = a
1 +ib
1
z
2 = a
2 +ib
2
z
3 = a
3 +ib
3
z
1 „ z
2 ⇔ a
1 < a
2
(a
1 = a
2
b
1 6 b
2
) z
2 „ z
3 ⇔ a
2 < a
3
(a
2 = a
3
b
2 6 b
3
) a
1 <a
2
a
1 <a
3
z
2 „z
3 ⇒a
26a
3
z
1 „z
3
a
1 = a
2
b
1 6 b
2
a
2 < a
3 ⇒ a
1 < a
3 ⇒ z
1 „ z
3
a
2 = a
3
b
26b
3 ⇒a
1 =a
3
b
16b
3 ⇒z
1 „z
3
z
1 =a
1 +ib
1
z
2 =a
2 +ib
2
a
1 <a
2
z
1 „z
2
a
1 =a
2
b
16b
2 ⇒z
1 „z
2
b
1 >b
2 ⇒z
2 „z
1
a
1 >a
2
z
2 „z
1
z
1
z
2
z
1 = a
1 +ib
1
z
2 = a
2 +ib
2
z
3 = a
3 +ib
3
z
1 „ z
2
a
1 < a
2
(a
1 = a
2
b
1 6 b
2
) R a
1 < a
2
(a
1 = a
2
b
1 6 b
2
) ⇔
a
1 +a
3 <a
2 +a
3
(a
1 +a
3 =a
2 +a
3
b
1 +b
36b
2 +b
3
)⇔z
1 +z
3 „z
2 +z
3
0 = 0+i0 i = 0+i 0 „ i 06 1 0×i = 0 i×i = i
2
= −1
i
2
„ 0 11 2007