Bonjour pouvez vous m'aider svp. Merci
Résoudre: 1/2x+3>7/10x-2
Factoriser: (x-3)(x-4)+(x-3)(5x+2)
Développer: (x-3)(x+2)-(2x+1)au carré
Résoudre: 1/2x+3>7/10x-2
Factoriser: (x-3)(x-4)+(x-3)(5x+2)
Développer: (x-3)(x+2)-(2x+1)au carré
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Bonsoir,
Résoudre: 1/2x+3>7/10x-2
3 + 2 > 7x/10 - x/2
5 > 7x/10 - 5x/10
5 > 2x/10
5 > x/5
x < 5 * 5
x < 25
x € ] - inf ; 25 [
Factoriser: (x-3)(x-4)+(x-3)(5x+2)
= (x - 3)(x - 4 + 5x + 2)
= (x - 3)(6x - 2)
= 2(x - 3)(3x - 1)
Développer: (x-3)(x+2)-(2x+1)^2
= x^2 + 2x - 3x - 6 - 4x^2 - 4x - 1
= -3x^2 - 5x - 7
Verified answer
Salut !
Résoudre: 1/2x+3>7/10x-2
= 3 + 2 > 7x/10 - x/2
= 5 > 7x/10 - 5x/10
= 5 > 2x/10
= 5 > x/5
= x < 5 * 5
= x < 25
Donc , x est inférieur à 25 .
Factoriser: (x-3)(x-4)+(x-3)(5x+2)
= (x - 3)(x - 4 + 5x + 2)
= (x - 3)(6x - 2)
= 2(x - 3)(3x - 1)
Développer: (x-3)(x+2)-(2x+1)^2
= x^2 + 2x - 3x - 6 - 4x^2 - 4x -1
= -3x^2 - 5x - 7