1. Polynômes, équations polynomiales 2. Autres types d`équations 3

Rx4−2x3+x2−2x+ 1 = 0
x+1
xxn+1 xn
xn−1
xn
x−1
x
n∈N
x9−12x8+x7+x6+ 5x5−x4+x3−7x2+x+ 1
n Pn
n2 cos nx =Pn(2 cos x)
xcos(xπ)
P P (x) =
x P (P(x)) = x
x−49
50 +x−50 49 =49 x−50 +50 x−49.
P, Q, R, S P (x)Q(y)+R(x)S(y)=1
x y
P(x)
xP (x−1) = (x−26)P(x)x.
px+ 3 −4√x−1 + px+ 8 −6√x−1=1
cos2x+ cos22x+ cos23x= 1
x1+x2=x2
3
x2+x3=x2
4
x3+x4=x2
5
x4+x5=x2
1
x5+x1=x2
2
P100
n=1
1
n(n+ 1) P100
n=1
1
n(n+ 1)(n+ 2)
n∈N∗1−x2
1=x2
1−x2
2=x31−x2 n−1=xn1−x2
n=x1
(xn)n>1|x1|<2014 3xn−xn−1=n
n > 1x2014 10−6
u0= 5 un+1 =un+1
un45 < u1000 <50

Rx4−2x3+x2−2x+ 1 = 0
x= 0 x2x2+x−2−
2(x+x−1) + 1 = 0 x2+x−2= (x+x−1)2−2y=x+x−1
y2−2y−1=0 y= 1 ±√2x2−yx + 1 = 0
x=1
2(y±py2−4) |y|>2
x=1
2(1 + √2±q−1+2√2).
x+1
xxn+1 xn
xn−1
xn
x−1
x
n∈N
Tn
n xn+x−n=Tn(x+x−1)
T0= 2 T1(X) = X(x+x−1)(xn+x−n) =
(xn+1 +x−n−1)+(xn−1+x−n+1)Tn+1(X) = XTn(X)−Tn−1(X)
x+1
xTn(x+x−1)
Un
nxn+1 −x−n−1
x−x−1=Un(x+x−1)
U0(X)=1 U1(X) = X
(x+x−1)(xn+1 −x−n−1) = (xn+2 −x−n−2)+(xn−x−n)
Un+1(X) = XUn(X)−Un−1(X)
x9−12x8+x7+x6+ 5x5−x4+x3−7x2+x+ 1
x=p/q p q
p9−12p8q+p7q2+p6q3+ 7p5q4−p4q5+p3q6−7p2q7+pq8+q9= 0.
p q9p=±1q p9q=±1
x=±1 1 −1
n Pn
n2 cos nx =Pn(2 cos x)
xcos(xπ)
2 cos nx cos x= cos(n+ 1)x+ cos(n−1)x
Pnn
x=m/n m n Tn(2 cos xπ) = 2 cos mπ =±2
2 cos xπ Tn∓2Tn
2 cos xπ cos xπ 0,±1
±1/2x1
2+k k k/3k

P P (x) = x
P(P(x)) = x
P(x)−x P (x)> x
x P (P(x)) > x x 
x−49
50 +x−50 49 =49 x−50 +50 x−49.
a= 49 b= 50
a(x−a)2(x−b) + b(x−a)(x−b)2=a2b(x−a) + ab2(x−b).
(x−a)(x−b)(a(x−a) + b(x−b)) = ab(a(x−a) + b(x−b))
x=a2+b2
a+b(x−a)(x−b) = ab
x2−(a+b)x= 0 x= 0 x=a+b x =a2+b2
a+b
P, Q, R, S P (x)Q(y) + R(x)S(y)=1
x y
P= 0 R(x)S(y) = 1 R S
Q R(x)S(y)y R S
P R 1
S(1 −QP (y))
P R
y z
S(y)S(z)
R(x) = 1
S(y)(1 −Q(y)P(x)) = 1 S(z)(1 −Q(z)P(x))
S(z)−S(y) + [S(y)Q(z)−S(z)Q(y)]P(x) = 0 S(y)Q(z)−S(z)Q(y)
P(x)S(y)Q(z)−S(z)Q(y)=0
S(z) = S(y)
S
S
P(x)
xP (x−1) = (x−26)P(x)x.
x= 0 0 P n
<26 P(n−1) = 0 0 = nP (n−1) = (n−26)P(n)P(n) = 0
0,1,...,25 P
P(x) = x(x−1) ···(x−25)Q(x).
Q(x) = Q(x−1)
Q(n) = Q(0) Q±∞ x
Q P (x) = ax(x−
1) ···(x−25) a∈R
px+ 3 −4√x−1 + px+ 8 −6√x−1=1
y=√x−1x=y2+1 p(y−2)2+p(y−3)2=
1|y−2|+|y−3|= 1
y>3 2y−5=1 y= 3

y62 2y−5 = −1y= 2
26y63 1 = 1
26y63 5 6x610 
cos2x+ cos22x+ cos23x= 1
2 cos2x= 1 + cos 2x
1 + cos 2x+ cos 4x+ cos 6x= 0.
1 + cos 6x= 2 cos23xcos 2x+ cos 4x= 2 cos xcos 3x
cos 3x(cos 3x+ cos x) = 0 cos xcos 2xcos 3x= 0
x= (k+1
2)π x = (k+1 2)π 2x= (k+1 2)π 3k∈Z
x1+x2=x2
3
x2+x3=x2
4
x3+x4=x2
5
x4+x5=x2
1
x5+x1=x2
2
x1=··· =x5= 2
i xix2
i=xi−1+xi−262xixi62
xj62
2Pxi=Px2
iP(xi−
1)2= 5 1
(xi−1)2= 1 i xi= 2 i
P100
n=1
1
n(n+ 1) P100
n=1
1
n(n+ 1)(n+ 2)
1
n(n+ 1) =1 n−1 n+ 1 n= 1,...,100
1−1
101 =100 101
a, b, c 1
n(n+ 1)(n+ 2) =a n+b n+ 1 +c n+ 2
1 = a(n+ 1)(n+ 2) + bn(n+ 2) + cn(n+ 1),
1 = (a+b+c)n2+ (3a+ 2b+c)n+ 2a a +b+c= 3a+ 2b+c= 0
2a= 1 a= 1/2b=−1c= 1/2
1
4−1 202 +1 204 
n∈N∗1−x2
1=x21−x2 2=
x31−x2
n−1=xn1−x2
n=x1
f(x)=1−x2xi+1 =f(xi)
16i6n xn+1 =x1
f f(x) = x x2+x−1 = 0
a=−1−√5
2b=−1 + √5 2

•xi=a i
•xi=b i
f(0) = 1 f(1) = 0 n
•xi= 0 i xi= 1 i
•xi= 1 i xi= 0 i
(xk)16k6n
a, b, 0,1
x1< a f(x)< x x < a x2=f(x1)< x1
x1> x2>··· > xn> x1
0 1
x7→ f(f(x)) −x0,1, a, b
f[0,1] [0,1] f◦f
[0,1] k xk[0,1]
xk6xk+2 f◦f xk+2 6xk+4
0,1, a b
xk+2 6xk
0 1
xka0a6f(x)61
x∈[a, 0] xk+1 ∈[a, 1] 0
1xk+1 ∈[a, 0]
f[a, 0]
61xk>1k
f[1,+∞[
(xn)n>1|x1|<2014 3xn−xn−1=n
n > 1x2014 10−6
xn=an +b
a b n = 3(an +b)−
a(n−1) −b= 2an + (2b+a)a= 1/2b=−1/4
yn=xn−(n
2−1 4)
3yn=yn−1|y1|<2015 |y2014|=
y1
32013 
<2015
32013 <10−6
1006,75 x2014 10−6
u0= 5 un+1 =un+1
un45 < u1000 <50
xn=u2
n
x0= 25 xn+1 =xn+1
xn+ 2 xn+1 >xn+ 2
xn>25 + 2n x1000 >2025 = 452
xn+1 −xn= 2 + 1
xn
62 + 1
25+2nn0 999
x1000 −25 62000 + ( 1
25 +1 27 +··· [phone] + 1000 25 = 2040 x1000 6
2065 <502