Sagot :
G=x*2-4-(x+2)(3x-5)2
1) G=2x-4-(x+2)(6x-10)
=2x-4-(x+2)6x-(x+2)10
=2x-4-(6x²+12x)-(10x+20)
=2x-4-6x²+12x-10x+20
=2x+12x-10x-4+20-6x²
=14x-10x-4+20-6x²
=4x-4+20-6x²
=4x+16-6x²
2) x*2-4
=2*x-2*2
=2(x-2)
3) 2x-4-(x + 2) (3 x - 5)
=2x−4−(x+2)(3x−5)2x-4-{{(x+2)(3x-5)}}
=2−4−((3−5)+2(3−5))
Ainsi le resultat est 32++6.
4)G=2x-4-(x+2)(6x-10)
pour x=2
4-4-(2+2)(6*2-10)
=4-4(-4)(12-10)
=4-4-48+40
=-48+40
=-8
Je crois que j ai reussi!