corrigé

A∆B= (A∪B)\(A∩B)=(A∪B)∩(A∩B)c,\,
= (A∪B)∩(Ac∪Bc), ,
= (A∩Ac)∪(A∩Bc)∪(B∩Ac)∪(B∩Bc),∩ ∪,
=∅ ∪ (A∩Bc)∪(B∩Ac)∪ ∅,
= (A∩Bc)∪(B∩Ac) = A∆B
∆P(E) (A, B, C)∈ P(E)A∆(B∆C)=(A∆B)∆C
A∆A= (A∪A)\(A∩A) = A\A=∅=A∆A
A∆∅= (A∪ ∅)\(A∩ ∅) = A\∅ =A=A∆∅
∆P(E)P(E)P(E)
A, B ∈ P(E)A∆B=B∆A∆
A∈ P(E)A∆∅=A∅∆A A∆A=∅
∆ (P(E),∆)
⇐=B=C=⇒A∆B=A∆C
=⇒A∆B=A∆C=⇒B=C A∆B=A∆C
B=C
BB⊂C x ∈B x ∈A x /∈A
x∈A x ∈A∩B x /∈A∆B A∆B=A∆C x /∈A∆C x ∈A
x∈C x C A A∆C
x /∈A x ∈B x ∈A∆B A∆B=A∆C x ∈A∆C x /∈A
x∈C
x∈C B ⊂C
BB C C ⊂B B =C
A∆B=A∩B A =∅A x ∈A
x∈B x ∈A x ∈A∩B x /∈A∆B
x /∈B x /∈A∩B x ∈A∆B
x A A =∅A B
B=∅
P=X4−5X3+7X2−5X+6 X2+1
−(X4+X2)X2−5X+6
−5X3+6X2−5X+6
−(−5X3−5X)
6X2+6
−(6X2+6)
0
P= (X2+ 1)Q Q =X2−5X+ 6 0
P P = (X2+ 1)(X2−5X+ 6)
X2+ 1 i−iC[X]X2+ 1 = (X−i)(X+i)R[X]X2+ 1
X2−5X+ 6 ∆ = 1 (5 −1)/2 = 2 (5 + 1)/2=3
R[X]C[X]X2−5X+ 6 = (X−2)(X−3)
P
C[X]P= (X−i)(X+i)(X−2)(X−3)
R[X]P= (X2+ 1)(X−2)(X−3)

RT
x∈Rx−x= 0 = 0 ×T k ∈Zk= 0 x−x=kT xRTx
RT
x, y ∈RxRTy k ∈Zx−y=kT
y−x=−kT = (−k)T k0∈Zk0=−k y−x=k0T yRTxRT
x, y, z ∈RxRTy yRTz k, k0∈Zx−y=kT
y−z=k0T
x−z= (x−y)+(y−z) = kT +k0T= (k+k0)T.
k00 ∈Zk00 =k+k0x−z=k00 T xRTzRT
x∈RyC(x)xR0xR0y
k∈Zx−y=k×0=0 x=y y x
y=xC(x)⊂ {x}xR0x x ∈ C(x){x}⊂C(x)
C(x) = {x}
x, y ∈Rcos(x) = cos(y) sin(x) = sin(y)x y
2π k ∈Zx=y+ 2kπ x −y=k(2π)
xSy⇐⇒ ∃k∈Z, x =y+ 2kπ ⇐⇒ xR2πy.
S=R2πS
a= 48 b= 35
a b b
a=b×1 + 13,
b= 13 ×2 + 9,
13 = 9 ×1 + 4,
9=4×2 + 1 ,
4 = 1 ×4.
13 = a−b,
9 = b−2×13 = b−2(a−b) = 3b−2a,
4 = 13 −9 = (a−b)−(3b−2a) = 3a−4b,
1 = 9 −4×2 = (3b−2a)−2(3a−4b) = 11b−8a .
a∧b= 1
−8a+ 11b= 1 (x0, y0)=(−8,11)
(x0, y0)
(x, y)
48x+ 35y= 1
48x0+ 35y0= 1
48(x−x0) + 35(y−y0)=0⇐⇒ 48(x−x0) = 35(y0−y).(?)
48(x−x0)k∈Z
x−x0= 35k x =x0+35k(?) 48× 35k= 35(y0−y)y=y0−48k
(x, y)k∈Zx=−8 + 35k y = 11 −48k
(x, y) (−8 + 35k, 11 −48k)k∈Z
48x+ 35y= 48(−8 + 35k) + 35(11 −48k) = (48 ×(−8) + 35 ×11) = 1,
(−8,11) (x, y)
{(−8 + 35k, 11 −48k), k ∈Z}
M(x, y, z)P−−→ AM ~u
−−→
AM.~u = 0
48(x+ 11) + 35(y−35) + 24(z+ 13) = 0.
P48x+ 35y+ 24z−385 = 0
P~u(48,35,24) P0z−16 = 0
~n(0,0,1) ~u ~n
D
48x+ 35y+ 24z−385 = 0
z−16 = 0 ⇐⇒ 48x+ 35y+ 24 ×16 −385 = 0 z−16 = 0 ⇐⇒ 48x+ 35y= 1 z= 16.

MD(x, y, z) (x, y, z)
z= 16 48x+ 35y= 1 (x, y)k∈Z
(x, y) = (−8 + 35k, 11 −48k)k[−100; 100]
k(x, y, z)=(−8 + 35k, 11 −48k, 16)
0 (−8,11,16)
1 (27,−37,16)
2 (62,−85,16)
−1 (−43,59,16)
D(−8,11,16) MD
(x, y, z)OM =px2+y2+z2
A=X2+X+ 1 j= exp(2iπ/3) = −1/2 + i√3/2¯ j
j A A(j) = 0 1 + j+j2= 0 j2=−1−j
j
aj +b=cj +d⇐⇒ a 1
2+i√3 2!+b=c 1 2+i√3 2!+d⇐⇒ a 2+b+i √3a 2!=c 2+d+i √3c 2!
⇐⇒ 
a
2+b=c 2+d
√3a
2=√3c 2
⇐⇒ (a=c
b=d,
P A R (R)<(A) (A) = 2 R
1R=αX +β α, β ∈RA P
P A P =AQ +R R =αX +β Q
j P (j) = A(j)Q(j) + R(j)A(j)=0 j A
P(j) = jn+jm+ 1 R(j) = αj +β jn+jm+ 1 = αj +β
n3Z0 1 2
n= 3k n = 3k+ 1 n= 3k+ 2 k
jn+jm+ 1 j3= 1 j2=−j−1
a b jn+jm+ 1 = aj +b
α β jn+jm+ 1 = αj +β
aj +b=αj +β α =a β =b
R=αX +β
n= 3k m = 3k0+ 2
jn+jm+ 1 = j3k+j3k0+2 + 1 = (j3)k+ (j3)k0×j2+ 1 = 1k+ 1k0×j2+ 1 = 1 + (−j−1) + 1 = −j+ 1,
αj +β=−j+ 1 α=−1β= 1 R=−X+ 1
n m jn+jm+ 1 α β R
3k3k03 0 3 3
3k3k0+ 1 j+ 2 1 2 X+ 2
3k3k0+ 2 j2+ 2 = −j+ 1 −1 1 −X+ 1
3k+ 1 3k0j+ 2 1 2 X+ 2
3k+ 1 3k0+ 1 2j+ 1 2 1 2X+ 1
3k+ 1 3k0+ 2 j2+j+ 1 = 0 0 0 0
3k+ 2 3k0j2+ 2 = −j+ 1 −1 1 −X+ 1
3k+ 2 3k0+ 1 j2+j+ 1 = 0 0 0 0
3k+ 2 3k0+ 2 2j2+ 1 = −2j−1−2−1−2X−1
A P R P A
R n = 3k+ 1 m= 3k0+ 2 n= 3k+ 2 m= 3k0+ 1
A|P⇐⇒ n≡1[3]
m≡2[3] n≡2[3] m≡1[3]