En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Sujet de brevetIdentités remarquables/Exercices/Sujet de brevet », n'a pu être restituée correctement ci-dessus.
Cet exercice est tombé au brevet, série collège (2000).
Soit D =
9
x
2
−
1
{\displaystyle 9x^{2}-1}
a) Quelle identité remarquable permet de factoriser D ?
Solution
D est sous la forme
a
2
−
b
2
{\displaystyle a^{2}-b^{2}}
b) Factoriser D.
Solution
D
=
9
x
2
−
1
{\displaystyle D=9x^{2}-1}
=
(
3
x
)
2
−
(
1
)
2
{\displaystyle =\left(3x\right)^{2}-\left(1\right)^{2}}
=
(
3
x
+
1
)
(
3
x
−
1
)
{\displaystyle =\left(3x+1\right)\left(3x-1\right)}
Soit
E
=
(
3
x
+
1
)
2
+
9
x
2
−
1
{\displaystyle E=\left(3x+1\right)^{2}+9x^{2}-1}
c) Développer E.
Solution
c)
E
=
(
3
x
+
1
)
2
+
9
x
2
−
1
{\displaystyle E=\left(3x+1\right)^{2}+9x^{2}-1}
=
(
9
x
2
+
2
×
3
x
×
1
+
1
)
+
9
x
2
−
1
{\displaystyle =\left(9x^{2}+2\times 3x\times 1+1\right)+9x^{2}-1}
=
9
x
2
+
6
x
+
1
+
9
x
2
−
1
{\displaystyle =9x^{2}+6x+1+9x^{2}-1}
=
18
x
2
+
6
x
{\displaystyle =18x^{2}+6x}
d) Factoriser E.
Solution
E
=
18
x
2
+
6
x
{\displaystyle E=18x^{2}+6x}
E
=
6
x
(
3
x
+
1
)
{\displaystyle E=6x\left(3x+1\right)}
e) Déterminer les solutions de l'équation
6
x
(
3
x
+
1
)
=
0
{\displaystyle 6x\left(3x+1\right)=0}
Solution
6
x
(
3
x
+
1
)
=
0
{\displaystyle 6x\left(3x+1\right)=0}
{
6
x
=
0
3
x
+
1
=
0
{\displaystyle {\begin{cases}6x=0\\3x+1=0\end{cases}}}
{
x
=
0
3
x
=
−
1
{\displaystyle {\begin{cases}x=0\\3x=-1\end{cases}}}
{
x
=
0
x
=
−
1
3
{\displaystyle {\begin{cases}x=0\\x={\frac {-1}{3}}\end{cases}}}
S
=
{
0
;
−
1
3
}
{\displaystyle S=\left\{0;{\frac {-1}{3}}\right\}}
On donne l’expression suivante :
K
(
x
)
=
(
5
x
−
3
)
2
+
6
(
5
x
−
3
)
{\displaystyle K\left(x\right)=\left(5x-3\right)^{2}+6\left(5x-3\right)}
Développer et réduire l’expression K(x).
Solution
K
(
x
)
=
(
5
x
−
3
)
2
+
6
(
5
x
−
3
)
{\displaystyle K\left(x\right)=\left(5x-3\right)^{2}+6\left(5x-3\right)}
=
(
25
x
2
−
2
×
5
x
×
3
+
9
)
+
6
(
5
x
−
3
)
{\displaystyle \left(25x^{2}-2\times 5x\times 3+9\right)+6\left(5x-3\right)}
=
25
x
2
−
30
x
+
9
+
30
x
−
18
{\displaystyle 25x^{2}-30x+9+30x-18}
=
25
x
2
−
9
{\displaystyle 25x^{2}-9}
Calculer
K
(
2
)
{\displaystyle K\left({\sqrt {2}}\right)}
Solution
K
(
2
)
=
25
(
2
)
2
−
9
{\displaystyle K\left({\sqrt {2}}\right)=25\left({\sqrt {2}}\right)^{2}-9}
K
(
2
)
=
25
×
2
−
9
{\displaystyle K\left({\sqrt {2}}\right)=25\times 2-9}
K
(
2
)
{\displaystyle K\left({\sqrt {2}}\right)}
Développer et réduire :
(
3
−
5
)
2
{\displaystyle \left(3-{\sqrt {5}}\right)^{2}}
Solution
(
3
−
5
)
2
{\displaystyle \left(3-{\sqrt {5}}\right)^{2}}
=
3
2
−
2
×
3
×
5
+
5
2
{\displaystyle 3^{2}-2\times 3\times {\sqrt {5}}+{\sqrt {5}}^{2}}
=
9
−
6
×
5
+
5
{\displaystyle 9-6\times {\sqrt {5}}+5}
=
14
−
6
×
5
{\displaystyle 14-6\times {\sqrt {5}}}
On considère l’expression :
E
=
(
x
−
3
)
2
−
(
x
−
1
)
(
x
−
2
)
{\displaystyle E=\left(x-3\right)^{2}-\left(x-1\right)\left(x-2\right)}
Développer et réduire E.
Solution
E
=
(
x
−
3
)
2
−
(
x
−
1
)
(
x
−
2
)
{\displaystyle E=\left(x-3\right)^{2}-\left(x-1\right)\left(x-2\right)}
=
(
x
2
−
2
×
3
×
x
+
9
)
−
(
x
2
−
2
x
−
x
+
2
)
{\displaystyle \left(x^{2}-2\times 3\times x+9\right)-\left(x^{2}-2x-x+2\right)}
=
x
2
−
6
x
+
9
−
x
2
+
2
x
+
x
−
2
{\displaystyle x^{2}-6x+9-x^{2}+2x+x-2}
=
−
3
x
+
7
{\displaystyle -3x+7}
Comment peut-on en déduire, sans calculatrice, le résultat de
99997
2
−
99999
×
99998
{\displaystyle 99997^{2}-99999\times 99998}
Solution
Il suffit de prendre x = 1000000
99997
2
−
99999
×
99998
=
−
3
×
1000000
+
7
{\displaystyle 99997^{2}-99999\times 99998=-3\times 1000000+7}
99997
2
−
99999
×
99998
=
−
2999993
{\displaystyle 99997^{2}-99999\times 99998=-2999993}
Factoriser l’expression :
F
=
(
4
x
+
1
)
2
−
(
4
x
+
1
)
(
7
x
−
6
)
{\displaystyle F=\left(4x+1\right)^{2}-\left(4x+1\right)\left(7x-6\right)}
Solution
F
=
(
4
x
+
1
)
2
−
(
4
x
+
1
)
(
7
x
−
6
)
{\displaystyle F=\left(4x+1\right)^{2}-\left(4x+1\right)\left(7x-6\right)}
=
[
4
x
+
1
]
[
(
4
x
+
1
)
−
(
7
x
−
6
)
]
{\displaystyle \left[4x+1\right]\left[\left(4x+1\right)-\left(7x-6\right)\right]}
Résoudre l'équation :
(
4
x
+
1
)
(
7
−
3
x
)
=
0
{\displaystyle \left(4x+1\right)\left(7-3x\right)=0}
Solution
(
4
x
+
1
)
(
7
−
3
x
)
=
0
{\displaystyle \left(4x+1\right)\left(7-3x\right)=0}
{
4
x
+
1
=
0
7
−
3
x
=
0
{\displaystyle {\begin{cases}4x+1=0\\7-3x=0\end{cases}}}
{
4
x
=
−
1
−
3
x
=
−
7
{\displaystyle {\begin{cases}4x=-1\\-3x=-7\end{cases}}}
{
x
=
−
1
4
x
=
7
3
{\displaystyle {\begin{cases}x=-{\frac {1}{4}}\\x={\frac {7}{3}}\end{cases}}}
Calculer :
(
4
−
5
)
2
{\displaystyle \left(4-{\sqrt {5}}\right)^{2}}
Solution
(
4
−
5
)
2
{\displaystyle \left(4-{\sqrt {5}}\right)^{2}}
=
4
2
−
2
×
4
×
5
+
5
2
{\displaystyle 4^{2}-2\times 4\times {\sqrt {5}}+{\sqrt {5}}^{2}}
=
16
−
8
5
+
5
{\displaystyle 16-8{\sqrt {5}}+5}
=
21
−
8
5
{\displaystyle 21-8{\sqrt {5}}}
On donne
F
=
(
4
x
−
3
)
2
−
(
x
+
3
)
(
3
−
9
x
)
{\displaystyle F=\left(4x-3\right)^{2}-\left(x+3\right)\left(3-9x\right)}
Développer et réduire
(
4
x
−
3
)
2
{\displaystyle \left(4x-3\right)^{2}}
Solution
(
4
x
−
3
)
2
{\displaystyle \left(4x-3\right)^{2}}
=
16
x
2
−
2
×
4
x
×
3
+
9
{\displaystyle 16x^{2}-2\times 4x\times 3+9}
=
16
x
2
−
24
x
+
9
{\displaystyle 16x^{2}-24x+9}
Montrer que
F
=
(
5
x
)
2
{\displaystyle F=\left(5x\right)^{2}}
Solution
F
=
(
4
x
−
3
)
2
−
(
x
+
3
)
(
3
−
9
x
)
{\displaystyle F=\left(4x-3\right)^{2}-\left(x+3\right)\left(3-9x\right)}
=
16
x
2
−
24
x
+
9
−
(
x
+
3
)
(
3
−
9
x
)
{\displaystyle 16x^{2}-24x+9-\left(x+3\right)\left(3-9x\right)}
=
16
x
2
−
24
x
+
9
−
(
3
x
−
9
x
2
+
9
−
27
x
)
{\displaystyle 16x^{2}-24x+9-\left(3x-9x^{2}+9-27x\right)}
=
16
x
2
−
24
x
+
9
−
3
x
+
9
x
2
−
9
+
27
x
{\displaystyle 16x^{2}-24x+9-3x+9x^{2}-9+27x}
=
25
x
2
{\displaystyle 25x^{2}}
=
(
5
x
)
2
{\displaystyle \left(5x\right)^{2}}
Trouvez les valeurs de x pour lesquelles F = 125
Solution
F
=
(
5
x
)
2
{\displaystyle F=\left(5x\right)^{2}}
(
5
x
)
2
=
125
{\displaystyle \left(5x\right)^{2}=125}
(
5
x
)
=
±
125
{\displaystyle \left(5x\right)=\pm {\sqrt {125}}}
(
5
x
)
=
±
25
×
5
{\displaystyle \left(5x\right)=\pm {\sqrt {25\times 5}}}
5
x
=
±
5
5
{\displaystyle 5x=\pm 5{\sqrt {5}}}
x
=
±
5
{\displaystyle x=\pm {\sqrt {5}}}
Soit l’expression
E
=
(
x
−
1
)
2
−
4
{\displaystyle E=\left(x-1\right)^{2}-4}
Calculer
E
{\displaystyle E}
x
=
0
{\displaystyle x=0}
Solution
E
=
(
x
−
1
)
2
−
4
{\displaystyle E=(x-1)^{2}-4}
E
=
(
−
1
)
2
−
4
{\displaystyle E=(-1)^{2}-4}
E
=
1
−
4
{\displaystyle E=1-4}
E
=
−
3
{\displaystyle E=-3}
Calculer la valeur exacte de
E
{\displaystyle E}
x
=
2
{\displaystyle x={\sqrt {2}}}
Solution
E
=
(
x
−
1
)
2
−
4
{\displaystyle E=\left(x-1\right)^{2}-4}
E
=
(
2
−
1
)
2
−
4
{\displaystyle E=\left({\sqrt {2}}-1\right)^{2}-4}
E
=
(
(
2
)
2
−
2
×
2
×
1
+
1
2
)
−
4
{\displaystyle E=\left(({\sqrt {2}})^{2}-2\times {\sqrt {2}}\times 1+1^{2}\right)-4}
E
=
(
2
−
2
2
+
1
)
−
4
{\displaystyle E=\left(2-2{\sqrt {2}}+1\right)-4}
E
=
−
1
−
2
2
{\displaystyle E=-1-2{\sqrt {2}}}
Factoriser
E
{\displaystyle E}
Solution
E
=
(
x
−
1
)
2
−
4
{\displaystyle E=(x-1)^{2}-4}
E
=
(
x
−
1
)
2
−
2
2
{\displaystyle E=(x-1)^{2}-2^{2}}
E
=
(
x
−
1
−
2
)
(
x
−
1
+
2
)
{\displaystyle E=(x-1-2)(x-1+2)}
E
=
(
x
−
3
)
(
x
+
1
)
{\displaystyle E=(x-3)(x+1)}
Résoudre l'équation :
(
x
+
1
)
(
x
−
3
)
=
0
{\displaystyle (x+1)(x-3)=0}
Solution
{
x
+
1
=
0
x
−
3
=
0
{\displaystyle {\begin{cases}x+1=0\\x-3=0\end{cases}}}
{
x
=
−
1
x
=
3
{\displaystyle {\begin{cases}x=-1\\x=3\end{cases}}}
On considère l’expression :
E
=
(
2
x
−
3
)
(
5
−
2
x
)
−
(
2
x
−
3
)
2
{\displaystyle E=(2x-3)(5-2x)-(2x-3)^{2}}
Développer et réduire E
Solution
E
=
(
2
x
−
3
)
(
5
−
2
x
)
−
(
2
x
−
3
)
2
{\displaystyle E=(2x-3)(5-2x)-(2x-3)^{2}}
=
(
10
x
−
4
x
2
−
15
+
6
x
)
−
(
4
x
2
−
12
x
+
9
)
{\displaystyle (10x-4x^{2}-15+6x)-(4x^{2}-12x+9)}
=
−
8
x
2
+
28
x
−
24
{\displaystyle -8x^{2}+28x-24}
Factoriser E.
Solution
E
=
(
2
x
−
3
)
(
5
−
2
x
)
−
(
2
x
−
3
)
2
{\displaystyle E=(2x-3)(5-2x)-(2x-3)2}
=
[
2
x
−
3
]
[
(
5
−
2
x
)
−
(
2
x
−
3
)
]
{\displaystyle \left[2x-3\right]\left[\left(5-2x\right)-\left(2x-3\right)\right]}
=
[
2
x
−
3
]
[
8
−
4
x
]
{\displaystyle \left[2x-3\right]\left[8-4x\right]}
Résoudre l'équation (2x - 3) (-4x + 8) = 0
Solution
{
2
x
−
3
=
0
−
4
x
+
8
=
0
{\displaystyle {\begin{cases}2x-3=0\\-4x+8=0\end{cases}}}
{
2
x
=
3
−
4
x
=
−
8
{\displaystyle {\begin{cases}2x=3\\-4x=-8\end{cases}}}
{
x
=
3
2
x
=
2
{\displaystyle {\begin{cases}x={\frac {3}{2}}\\x=2\end{cases}}}
On donne l’expression suivante :
F
=
(
2
x
+
3
)
2
−
(
x
+
5
)
(
2
x
+
3
)
{\displaystyle F=(2x+3)^{2}-(x+5)(2x+3)}
Développer et réduire.
Solution
F
=
(
2
x
+
3
)
2
−
(
x
+
5
)
(
2
x
+
3
)
{\displaystyle F=(2x+3)^{2}-(x+5)(2x+3)}
=
(
4
x
2
+
12
x
+
9
)
−
(
2
x
2
+
3
x
+
10
x
+
15
)
{\displaystyle (4x^{2}+12x+9)-(2x^{2}+3x+10x+15)}
=
2
x
2
−
x
−
6
{\displaystyle 2x^{2}-x-6}
Factoriser
Solution
F
=
(
2
x
+
3
)
2
−
(
x
+
5
)
(
2
x
+
3
)
{\displaystyle F=(2x+3)^{2}-(x+5)(2x+3)}
=
[
2
x
+
3
]
[
(
2
x
+
3
)
−
(
x
+
5
)
]
{\displaystyle \left[2x+3\right]\left[(2x+3)-(x+5)\right]}
=
[
2
x
+
3
]
[
x
−
2
]
]
{\displaystyle \left[2x+3\right]\left[x-2]\right]}
Résoudre l'équation (2x + 3)(x-2) = 0.
Solution
{
2
x
+
3
=
0
x
−
2
=
0
{\displaystyle {\begin{cases}2x+3=0\\x-2=0\end{cases}}}
{
2
x
=
−
3
x
=
2
{\displaystyle {\begin{cases}2x=-3\\x=2\end{cases}}}
{
x
=
−
3
2
x
=
2
{\displaystyle {\begin{cases}x=-{\frac {3}{2}}\\x=2\end{cases}}}
On pose
E
=
(
5
x
−
2
)
(
x
+
7
)
+
(
5
x
−
2
)
2
{\displaystyle E=(5x-2)(x+7)+(5x-2)^{2}}
Développer et réduire E.
Solution
E
=
(
5
x
−
2
)
(
x
+
7
)
+
(
5
x
−
2
)
2
E
=
(
5
x
2
+
35
x
−
2
x
−
14
)
+
(
25
x
2
−
20
x
+
4
)
E
=
30
x
2
+
13
x
−
10
{\displaystyle {\begin{aligned}&E=(5x-2)(x+7)+(5x-2)^{2}\\&E=(5x^{2}+35x-2x-14)+(25x^{2}-20x+4)\\&E=30x^{2}+13x-10\\\end{aligned}}}
Factoriser E.
Solution
E
=
(
5
x
−
2
)
(
x
+
7
)
+
(
5
x
−
2
)
2
E
=
(
5
x
−
2
)
(
(
x
+
7
)
+
(
5
x
−
2
)
)
E
=
(
5
x
−
2
)
(
6
x
+
5
)
{\displaystyle {\begin{aligned}&E=(5x-2)(x+7)+(5x-2)^{2}\\&E=(5x-2)((x+7)+(5x-2))\\&E=(5x-2)(6x+5)\\\end{aligned}}}
Calculer E pour
x
=
2
5
{\displaystyle x={\frac {2}{5}}}
Solution
E
=
(
5
x
−
2
)
(
6
x
+
5
)
E
=
(
5
×
2
5
−
2
)
(
6
×
2
5
+
5
)
E
=
0
×
(
12
5
+
5
)
E
=
0
{\displaystyle {\begin{aligned}&E=(5x-2)(6x+5)\\&E=(5\times {\frac {2}{5}}-2)(6\times {\frac {2}{5}}+5)\\&E=0\times ({\frac {12}{5}}+5)\\&E=0\\\end{aligned}}}
Résoudre l'équation
(
5
x
−
2
)
(
6
x
+
5
)
=
0
{\displaystyle (5x-2)(6x+5)=0}
Solution
{
5
x
−
2
=
0
6
x
+
5
=
0
{\displaystyle {\begin{cases}5x-2=0\\6x+5=0\end{cases}}}
{
5
x
=
2
6
x
=
−
5
{\displaystyle {\begin{cases}5x=2\\6x=-5\end{cases}}}
{
x
=
2
5
x
=
−
5
6
{\displaystyle {\begin{cases}x={\frac {2}{5}}\\x={\frac {-5}{6}}\end{cases}}}
S
=
{
2
5
;
−
5
6
}
{\displaystyle S=\{{\frac {2}{5}};{\frac {-5}{6}}\}}
Développer en utilisant les identités remarquables, puis simplifier.
a)
(
2
x
+
3
)
2
+
5
(
x
−
7
)
2
{\displaystyle (2x+3)^{2}+5(x-7)^{2}}
Solution
A
=
(
2
x
+
3
)
2
+
5
(
x
−
7
)
2
A
=
(
4
x
2
+
12
x
+
9
)
+
5
(
x
2
−
14
x
+
49
)
A
=
4
x
2
+
12
x
+
9
+
5
x
2
−
70
x
+
245
)
A
=
9
x
2
−
58
x
+
254
{\displaystyle {\begin{aligned}&A=(2x+3)^{2}+5(x-7)^{2}\\&A=(4x^{2}+12x+9)+5(x^{2}-14x+49)\\&A=4x^{2}+12x+9+5x^{2}-70x+245)\\&A=9x^{2}-58x+254\\\end{aligned}}}
b)
(
3
x
−
5
)
2
−
(
2
x
−
3
)
(
2
x
+
3
)
{\displaystyle (3x-5)^{2}-(2x-3)(2x+3)}
Solution
B
=
(
3
x
−
5
)
2
−
(
2
x
−
3
)
(
2
x
+
3
)
B
=
(
9
x
2
−
30
x
+
25
)
−
(
4
x
2
−
9
)
B
=
5
x
2
−
30
x
+
34
{\displaystyle {\begin{aligned}&B=(3x-5)^{2}-(2x-3)(2x+3)\\&B=(9x^{2}-30x+25)-(4x^{2}-9)\\&B=5x^{2}-30x+34\\\end{aligned}}}
Développer en utilisant les identités remarquables, puis simplifier.
a)
(
2
x
+
7
)
2
+
5
(
x
−
3
)
2
{\displaystyle (2x+7)^{2}+5(x-3)^{2}}
Solution
A
=
(
2
x
+
7
)
2
+
5
(
x
−
3
)
2
A
=
(
4
x
2
+
28
x
+
49
)
+
5
(
x
2
−
6
x
+
9
)
A
=
4
x
2
+
28
x
+
49
+
5
x
2
−
30
x
+
45
A
=
9
x
2
−
2
x
+
94
{\displaystyle {\begin{aligned}&A=(2x+7)^{2}+5(x-3)^{2}\\&A=(4x^{2}+28x+49)+5(x^{2}-6x+9)\\&A=4x^{2}+28x+49+5x^{2}-30x+45\\&A=9x^{2}-2x+94\\\end{aligned}}}
b)
(
5
x
−
3
)
2
−
(
3
x
−
2
)
(
3
x
+
2
)
{\displaystyle (5x-3)^{2}-(3x-2)(3x+2)}
Solution
B
=
(
5
x
−
3
)
2
−
(
3
x
−
2
)
(
3
x
+
2
)
B
=
(
25
x
2
−
30
x
+
9
)
−
(
9
x
2
−
4
)
B
=
16
x
2
−
30
x
+
13
{\displaystyle {\begin{aligned}&B=(5x-3)^{2}-(3x-2)(3x+2)\\&B=(25x^{2}-30x+9)-(9x^{2}-4)\\&B=16x^{2}-30x+13\\\end{aligned}}}
Cet exercice est tombé au brevet des collèges (1995).
Soit P=
(
x
−
2
)
(
2
x
+
1
)
−
(
2
x
+
1
)
2
{\displaystyle (x-2)(2x+1)-(2x+1)^{2}}
a) Développer et réduire l’expression P.
Solution
P=
(
x
−
2
)
(
2
x
+
1
)
−
(
2
x
+
1
)
2
{\displaystyle (x-2)(2x+1)-(2x+1)^{2}}
P=
2
x
2
+
x
−
4
x
−
2
−
(
4
x
2
+
4
x
+
1
)
{\displaystyle 2x^{2}+x-4x-2-(4x^{2}+4x+1)}
P=
2
x
2
+
x
−
4
x
−
2
−
4
x
2
−
4
x
−
1
)
{\displaystyle 2x^{2}+x-4x-2-4x^{2}-4x-1)}
P=
−
2
x
2
−
7
x
−
3
{\displaystyle -2x^{2}-7x-3}
b) Factoriser P.
Solution
P=
(
x
−
2
)
(
2
x
+
1
)
−
(
2
x
+
1
)
2
{\displaystyle (x-2)(2x+1)-(2x+1)^{2}}
P=
(
2
x
+
1
)
[
(
x
−
2
)
−
(
2
x
+
1
)
]
{\displaystyle (2x+1)\left[(x-2)-(2x+1)\right]}
P=
(
2
x
+
1
)
[
−
x
−
3
]
{\displaystyle (2x+1)\left[-x-3\right]}
P=
(
2
x
+
1
)
(
−
x
−
3
)
{\displaystyle (2x+1)(-x-3)}
c) Résoudre l'équation
(
2
x
+
1
)
(
x
+
3
)
=
0
{\displaystyle (2x+1)(x+3)=0}
Solution
{
2
x
+
1
=
0
x
+
3
=
0
{\displaystyle {\begin{cases}2x+1=0\\x+3=0\end{cases}}}
{
x
=
−
1
2
x
=
−
3
{\displaystyle {\begin{cases}x=-{\frac {1}{2}}\\x=-3\end{cases}}}
d) Pour
x
=
−
3
7
{\displaystyle x=-{\frac {3}{7}}}
Solution
P=
(
x
−
2
)
(
2
x
+
1
)
−
(
2
x
+
1
)
2
{\displaystyle (x-2)(2x+1)-(2x+1)^{2}}
P=
(
−
3
7
−
2
)
(
2
−
3
7
+
1
)
−
(
2
−
3
7
+
1
)
2
{\displaystyle ({\frac {-3}{7}}-2)(2{\frac {-3}{7}}+1)-(2{\frac {-3}{7}}+1)^{2}}
P=
(
−
3
7
−
14
7
)
(
−
6
7
+
7
7
)
−
(
−
6
7
+
7
7
)
2
{\displaystyle (-{\frac {3}{7}}-{\frac {14}{7}})(-{\frac {6}{7}}+{\frac {7}{7}})-({\frac {-6}{7}}+{\frac {7}{7}})^{2}}
P=
(
−
17
7
)
(
1
7
)
−
(
1
7
)
2
{\displaystyle (-{\frac {17}{7}})({\frac {1}{7}})-({\frac {1}{7}})^{2}}
P=
−
17
×
1
7
×
7
−
1
2
7
2
{\displaystyle -{\frac {17\times 1}{7\times 7}}-{\frac {1^{2}}{7^{2}}}}
P=
−
−
17
49
+
−
1
49
{\displaystyle -{\frac {-17}{49}}+{\frac {-1}{49}}}
P=
−
18
49
{\displaystyle {\frac {-18}{49}}}
Cet exercice est tombé au brevet des collèges (1995).
Soir l’expression F =
(
2
x
−
5
)
2
−
x
(
2
x
−
5
)
{\displaystyle (2x-5)^{2}-x(2x-5)}
a) Développer et réduire F.
Solution
F =
(
2
x
−
5
)
2
−
x
(
2
x
−
5
)
{\displaystyle (2x-5)^{2}-x(2x-5)}
F =
4
x
2
−
20
x
+
25
−
2
x
2
+
5
x
{\displaystyle 4x^{2}-20x+25-2x^{2}+5x}
F =
2
x
2
−
15
x
+
25
{\displaystyle 2x^{2}-15x+25}
b) Factoriser F.
Solution
F =
(
2
x
−
5
)
2
−
x
(
2
x
−
5
)
{\displaystyle (2x-5)^{2}-x(2x-5)}
F =
(
2
x
−
5
)
[
(
2
x
−
5
)
−
x
]
{\displaystyle (2x-5)\left[(2x-5)-x\right]}
F =
(
2
x
−
5
)
[
x
−
5
]
{\displaystyle (2x-5)\left[x-5\right]}
F =
(
2
x
−
5
)
(
x
−
5
)
{\displaystyle (2x-5)(x-5)}
c) Résoudre l'équation
(
2
x
−
5
)
(
x
−
5
)
=
0
{\displaystyle (2x-5)(x-5)=0}
Solution
{
2
x
−
5
=
0
x
−
5
=
0
{\displaystyle {\begin{cases}2x-5=0\\x-5=0\end{cases}}}
{
x
=
5
2
x
=
5
{\displaystyle {\begin{cases}x={\frac {5}{2}}\\x=5\end{cases}}}
![{\displaystyle \left[4x+1\right]\left[\left(4x+1\right)-\left(7x-6\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfc6b117e27f76f1e135f59c8f0692c8ab1dd86e)
![{\displaystyle \left[2x-3\right]\left[\left(5-2x\right)-\left(2x-3\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/596276b38b09ff10488a34ec1af1a1e84973db1a)
![{\displaystyle \left[2x-3\right]\left[8-4x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3ca25ddfaf9f8f1f03769562422ba057dd96341)
![{\displaystyle \left[2x+3\right]\left[(2x+3)-(x+5)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4a18876f131c2a7fe3fff3bbd64b213ba90b73f)
![{\displaystyle \left[2x+3\right]\left[x-2]\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7aa99432df29a1f3d1bab67c0abaeb4644ed2477)
![{\displaystyle (2x+1)\left[(x-2)-(2x+1)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea5219774b19a801589f6988759d30606cf5ca45)
![{\displaystyle (2x+1)\left[-x-3\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f53c904bab542376b6f5042d2846572bb0fbb7a)
![{\displaystyle (2x-5)\left[(2x-5)-x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb80522c30419091c0c30649bbe160e935e720b7)
![{\displaystyle (2x-5)\left[x-5\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff9f286721c541a096f734c564ed0fb773f6b55)
