Bonjour, je suis en seconde et j'aurais besoin d'aide pour cet exercice de maths !

Réponse:

A.

1)

g(x) = (x+2)(3-5x)-(3x-2)(-2x-4)

g(x) = (x+2)(3-5x)-(3x-2)[-2(x+2)]

g(x) = (x+2)(3-5x-(-2)(3x-2)]

g(x) = (x+2)(3-5x+6x-4)

g(x) = (x+2)(x-1)

2)

g(x) = (x+2)(x-1)

g(x) = x² - x + 2x - 2

g(x) = x² + x - 2

3)

[tex] {(x + \frac{1}{2} )}^{2} - \frac{9}{4} = \\ {x}^{2} + 2 \times x \times \frac{1}{2} + {( \frac{1}{2} )}^{2} - \frac{9}{4} = \\ {x}^{2} + x + \frac{1}{4} - \frac{9}{4} = \\ {x}^{2} + x - \frac{8}{4} = \\ {x}^{2} + x - 2 = \\ \: g(x) \\ [/tex]

B.

1.

[tex]g( - \frac{1}{2} ) = {( - \frac{1}{2} + \frac{1}{2}) }^{2} - \frac{9}{4} = - \frac{9}{4} [/tex]

[tex]g( \sqrt{2} ) = { \sqrt{2} }^{2} + \sqrt{2} - 2 = 2 + \sqrt{2} - 2 = \sqrt{2} [/tex]

2.

g(x)=0

(x+2)(x-1)=0

x+2 = 0 ou x-1=0

x= -2 ou x= 1

S ={-2; 1}

[tex]g(x) = - \frac{9}{4} \\ {(x + \frac{1}{2}) }^{2} - \frac{9}{4} = - \frac{9}{4} \\ {(x + \frac{1}{2}) }^{2} = 0 \\ x + \frac{1}{2} = 0 \\ x = - \frac{1}{2} \\ [/tex]

S ={-½}

g(x)=-2

x²+x-2=-2

x²+x=0

x(x+1)=0

x = 0 ou x+1=0

x= 0 ou x=-1

S = {-1; 0}