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}
