bonjour j'aurez bsoin d'aide pour un exercice pourriez vous m'aider s'il vous plait On considère A(x)=(2x+3)²-(7-3x)² 1)développer 2)résoudre dans r l'équation

Bonjour,

1) A(x) = (2x + 3)^2 - (7 - 3x)^2
A(x) = 4x^2 + 12x + 9 - 49 + 42x - 9x^2
A(x) = -5x^2 + 54x - 40

2) A(x) = -40

-5x^2 + 54x - 40 = -40
-5x^2 + 54x = 0
x(-5x + 54) = 0

x = 0
Ou
-5x + 54 = 0
5x = 54
x = 54 / 5
x = 10,8

3) A(x) = (2x + 3)^2 - (7 - 3x)^2
A(x) = (2x + 3 - 7 + 3x)(2x + 3 + 7 - 3x)
A(x) = (5x - 4)(-x + 10)

4) A(x) = 0

5x - 4 = 0
5x = 4
x = 4/5

-x + 10 = 0
x = 10

5) A(x) >> 0

(5x - 4)(-x + 10)

x..................|...-inf...........(4/5)..........10...........+inf
5x - 4.........|............(-).......O.....(+)..........(+)...........
-x + 10........|............(+)...............(+)....O.....(-).........
A(x)............|............(-).......O.......(+)....O....(-)..........

A(x) << 0 pour x € ]-inf ; 4/5] U [10 ; + inf [

Bonjour,

1) 

[tex]A(x) = (2x+3)^2-(7-3x)^2\\ \Rightarrow A(x) = (a+b)^2-(a-b)^2 = (a^2+2ab+b^2) - (a^2-2ab+b^2)\\ \Rightarrow A(x) = ((2x)^2+2\times2x\times3+3^2)-(7^2-2\times7\times3x+(3x)^2)\\ \Rightarrow A(x) = (4x^2+12x+9)-(49-42x+9x^2)\\ \Rightarrow A(x) =4x^2+12x+9-49+42x-9x^2\\ \Rightarrow A(x) = -5x^2+54x-40[/tex]

2) 

[tex]A(x) = -5x^2+54x-40 = -40\\\\ -5x^2+54x-40+40=0\\ -5x^2+54x=0\\ -x(5x-54)=0\\ -x=0\ \text{ ou } \ 5x-54=0\\ x=0 \ \text{ ou } \ x = \dfrac{54}{5}\\\\ \boxed{S = \left\{0; \dfrac{54}{5}\right\}}[/tex]

3) 

[tex]A(x) = -5x^2+54x-40\\ -5x^2+50x+4x-40\\ -5x(x-10)+4(x-10)\\ (-5x+4)(x-10)[/tex]

4) 

[tex]A(x) = (-5x+4)(x-10) = 0\\\\ -5x+4 = 0 \ \text{ ou } \ x-10=0\\\\ x = \dfrac{4}{5} \ \text{ ou } \ x = 10\\\\ \boxed{S = \left\{\dfrac{4}{5}; 10\right\}}[/tex]

5)

[tex]A(x) = (-5x+4)(x-10) \geq 0\\ \begin{cases}-5x+4\geq 0\\x-10 \geq 0\end{cases}\\\\ \begin{cases}-5x+4\leq 0\\x-10 \leq 0\end{cases}\\\\ \begin{cases}x\leq \dfrac{4}{5}\\ x \geq 10\end{cases} \Rightarrow x \in \varnothing\\\\ \begin{cases}x\geq \dfrac{4}{5}\\ x \leq 10\end{cases} \Rightarrow x \in \left[\dfrac{4}{5}, 10\right]\\\\\\ \boxed{S = x\in \left[\dfrac{4}{5}, 10\right]}[/tex]


Bonne journée.