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Calcul littéral : développement, factorisation, identités remarquables
Factorisation en utilisant les identités remarquables - Exercice 1
20 min
40
Soit
x
x
x
un réel, factoriser les expressions suivantes :
Question 1
A
=
x
2
+
2
x
+
1
A=x^{2} +2x+1
A
=
x
2
+
2
x
+
1
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
+
2
×
a
×
b
+
b
2
=
(
a
+
b
)
2
{\color{blue}a}^{2} +2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}+{\color{red}b}\right)^{2}
a
2
+
2
×
a
×
b
+
b
2
=
(
a
+
b
)
2
A
=
x
2
+
2
x
+
1
A=x^{2} +2x+1
A
=
x
2
+
2
x
+
1
A
=
x
2
+
2
×
x
×
1
+
1
2
A={\color{blue}x}^{2} +2\times {\color{blue}x}\times {\color{red}1}+{\color{red}1}^{2}
A
=
x
2
+
2
×
x
×
1
+
1
2
Ici nous avons
a
=
x
a={\color{blue}x}
a
=
x
et
b
=
1
b={\color{red}1}
b
=
1
. Il vient alors que :
A
=
(
x
+
1
)
2
A=\left({\color{blue}x}+{\color{red}1}\right)^{2}
A
=
(
x
+
1
)
2
Question 2
B
=
4
x
2
−
12
x
+
9
B=4x^{2} -12x+9
B
=
4
x
2
−
12
x
+
9
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
−
2
×
a
×
b
+
b
2
=
(
a
−
b
)
2
{\color{blue}a}^{2} -2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)^{2}
a
2
−
2
×
a
×
b
+
b
2
=
(
a
−
b
)
2
B
=
4
x
2
−
12
x
+
9
B=4x^{2} -12x+9
B
=
4
x
2
−
12
x
+
9
B
=
(
2
x
)
2
−
2
×
2
x
×
3
+
3
2
B=\left({\color{blue}2x}\right)^{2} -2\times{\color{blue}2x}\times {\color{red}3}+{\color{red}3}^{2}
B
=
(
2
x
)
2
−
2
×
2
x
×
3
+
3
2
Ici nous avons
a
=
2
x
a={\color{blue}2x}
a
=
2
x
et
b
=
3
b={\color{red}3}
b
=
3
. Il vient alors que :
B
=
(
2
x
−
3
)
2
B=\left({\color{blue}2x}-{\color{red}3}\right)^{2}
B
=
(
2
x
−
3
)
2
Question 3
C
=
25
x
2
−
49
C=25x^{2}-49
C
=
25
x
2
−
49
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
{\color{blue}a}^{2} -{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)\left({\color{blue}a}+{\color{red}b}\right)
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
C
=
25
x
2
−
49
C=25x^{2}-49
C
=
25
x
2
−
49
équivaut successivement à :
C
=
(
5
x
)
2
−
7
2
C=\left({\color{blue}5x}\right)^{2} -{\color{red}7}^{2}
C
=
(
5
x
)
2
−
7
2
Ici nous avons
a
=
5
x
a={\color{blue}5x}
a
=
5
x
et
b
=
7
b={\color{red}7}
b
=
7
. Il vient alors que :
C
=
(
5
x
−
7
)
(
5
x
+
7
)
C=\left({\color{blue}5x}-{\color{red}7}\right)\left({\color{blue}5x}+{\color{red}7}\right)
C
=
(
5
x
−
7
)
(
5
x
+
7
)
Question 4
D
=
81
−
64
x
2
D=81-64x^{2}
D
=
81
−
64
x
2
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
{\color{blue}a}^{2} -{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)\left({\color{blue}a}+{\color{red}b}\right)
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
D
=
81
−
64
x
2
D=81-64x^{2}
D
=
81
−
64
x
2
équivaut successivement à :
D
=
9
2
−
(
8
x
)
2
D={\color{blue}9}^{2}-\left({\color{red}8x}\right)^{2}
D
=
9
2
−
(
8
x
)
2
Ici nous avons
a
=
9
a={\color{blue}9}
a
=
9
et
b
=
8
x
b={\color{red}8x}
b
=
8
x
. Il vient alors que :
D
=
(
9
−
8
x
)
(
9
+
8
x
)
D=\left({\color{blue}9}-{\color{red}8x}\right)\left({\color{blue}9}+{\color{red}8x}\right)
D
=
(
9
−
8
x
)
(
9
+
8
x
)
Question 5
E
=
4
x
2
−
2
x
+
1
4
E=4x^{2} -2x+\frac{1}{4}
E
=
4
x
2
−
2
x
+
4
1
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
−
2
×
a
×
b
+
b
2
=
(
a
−
b
)
2
{\color{blue}a}^{2} -2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)^{2}
a
2
−
2
×
a
×
b
+
b
2
=
(
a
−
b
)
2
E
=
4
x
2
−
2
x
+
1
4
E=4x^{2} -2x+\frac{1}{4}
E
=
4
x
2
−
2
x
+
4
1
E
=
(
2
x
)
2
−
2
×
2
x
×
1
2
+
(
1
2
)
2
E=\left({\color{blue}2x}\right)^{2} -2\times {\color{blue}2x}\times {\color{red}\frac{1}{2}} +\left({\color{red}\frac{1}{2}} \right)^{2}
E
=
(
2
x
)
2
−
2
×
2
x
×
2
1
+
(
2
1
)
2
Ici nous avons
a
=
2
x
a={\color{blue}2x}
a
=
2
x
et
b
=
1
2
b={\color{red}\frac{1}{2}}
b
=
2
1
. Il vient alors que :
E
=
(
2
x
−
1
2
)
2
E=\left({\color{blue}2x}-{\color{red}\frac{1}{2}} \right)^{2}
E
=
(
2
x
−
2
1
)
2
Question 6
F
=
x
2
+
24
x
+
144
F=x^{2} +24x+144
F
=
x
2
+
24
x
+
144
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
+
2
×
a
×
b
+
b
2
=
(
a
+
b
)
2
{\color{blue}a}^{2} +2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}+{\color{red}b}\right)^{2}
a
2
+
2
×
a
×
b
+
b
2
=
(
a
+
b
)
2
F
=
x
2
+
24
x
+
144
F=x^{2} +24x+144
F
=
x
2
+
24
x
+
144
F
=
x
2
+
2
×
x
×
12
+
12
2
F={\color{blue}x}^{2} +2\times {\color{blue}x}\times {\color{red}12}+{\color{red}12}^{2}
F
=
x
2
+
2
×
x
×
12
+
12
2
Ici nous avons
a
=
x
a={\color{blue}x}
a
=
x
et
b
=
12
b={\color{red}12}
b
=
12
. Il vient alors que :
F
=
(
x
+
12
)
2
F=\left({\color{blue}x}+{\color{red}12}\right)^{2}
F
=
(
x
+
12
)
2
Question 7
G
=
64
x
2
−
80
x
+
25
G=64x^{2} -80x+25
G
=
64
x
2
−
80
x
+
25
Correction
Identit
e
ˊ
remarquable
\purple{\text{Identité remarquable}}
Identit
e
ˊ
remarquable
a
2
−
2
×
a
×
b
+
b
2
=
(
a
−
b
)
2
{\color{blue}a}^{2} -2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)^{2}
a
2
−
2
×
a
×
b
+
b
2
=
(
a
−
b
)
2
G
=
64
x
2
−
80
x
+
25
G=64x^{2} -80x+25
G
=
64
x
2
−
80
x
+
25
G
=
(
8
x
)
2
−
2
×
8
x
×
5
+
5
2
G=\left({\color{blue}8x}\right)^{2} -2\times{\color{blue}8x}\times {\color{red}5}+{\color{red}5}^{2}
G
=
(
8
x
)
2
−
2
×
8
x
×
5
+
5
2
Ici nous avons
a
=
8
x
a={\color{blue}8x}
a
=
8
x
et
b
=
5
b={\color{red}5}
b
=
5
. Il vient alors que :
G
=
(
8
x
−
5
)
2
G=\left({\color{blue}8x}-{\color{red}5}\right)^{2}
G
=
(
8
x
−
5
)
2