Calcul algébrique

•(a+b)3=
a3+ 3a2b+ 3ab2+b3
•(2x−4)(x+ 3) −(5 −x)(2 −x) = (x−2)(2(x+3)+5−x) = (x−2)(x+ 8)
•63×5
10√8=23×33×5 2×5×2√2=2×27
√2=
27√2
•(x−1)(2x+5) = 0
S=−5
2; 1
•0
x
ax +b a b
a
0a
(2x−4)(x+ 3)
5−x≤0
x−3 2 5
2x−4− − + +
x+ 3 −+ + +
5−x+ + + −
Q+−+−
QS= [−3; 2]∪]5; +∞[
•
•

•x0
•|x−2| ≥ 3x2 3
x∈]− ∞;−1] ∪[5; +∞[
• |a|=|b|a b
|2x−5|=|3−x| ⇔ 2x−5 = 3 −x2x−5 = −3 + x x =8
3x= 2
(2x+ 5)3
(2x+ 1)(x2−1) −3(x+ 1)(2x+1)+5x(2x+ 1)(x+ 1)
(x2+x−3)(2x+ 1)
3x(x2−6x+ 9) + 4(x−2)(x2−3x) + 7(x−3)x2
(4x2−9) + 2x+ 3
1 + √3−3√2
3−√2
a5(a2b)−3
(a−1b3)3b−2
(4−3×103)−2×27√80
68×√54
([−1; 0] ∪[2; 7[) ∩(] −2; 3]∪]6; +∞[)
3744
7920
251 ×249
4−x
2x+ 6 ≤1
(6 −4x)26(3x−5)2
|x−2|<4
x2+ 4x+ 4 = (3x+ 6)(x−3)
(x+ 1)2−3(x+ 1)(x−2) ≤0
|4x+ 1|=|−7x+ 3|x3x
2 5
x(2x−3)
(3 −x)(x+ 1) <0
|x−1|+|x+ 3|= 4
6x−x2−9>2 + |x+ 5|

x y z x ∈[1; 3] 2 < y < 3z2≤1
z
3x+ 2y x2+ 4x3
y−1
x2−2x+ 1
2z−3xy + 1 x−5
2z−3z(x+ 2) x y−3z2
738 7.8
310−2
11.5 1 3.14 ≤π≤3.15
x2−4x+ 3 = 0
x2−4x+ 4 = 1
x2+ 6x+ 5 = 0 4x2−4x+ 6 = 0
x A(x) = |x−1|+|2x−4| − |3 + x|
A(x) = 2
A(x)≤4
A(x) = 2x−1
2 000 10 2 000
8% 6%
4%
2 000
1 2 001
2 002 2 003
n
1 2 000 2 010 2 020 2 050
10 1 2 000
75

2 001 2 002
2 000 2 010 2 020 2 050
n+ 3 n n
2 45
p(p+ 1)2−p2= 2p+ 1
n
n
80
30
1
1+3+5+··· + 199
95
711 711
711 000