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Dianebtz
@Dianebtz
April 2019
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Voila j'ai un de mes exercices de math qui me pose problème, pouvez vous m'aidez s.v.p ?
factoriser les expressions puis résoudre les équations :
a. (t+1)(2t-5)+t(t+1)=0
b. 100t²+20t+1-(10t+1)x(t+5)=0
c. (t+3)(3t-6)+t²-9=0
d. x(x+4)+x²+8x+16=0
e. 49-x²-(7+x)(2x-3)=0
f.-5y(y-3)+9-6y+y²=0
Merci d'avance pour celui qui répondra !
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isapaul
Verified answer
Bonsoir
a) (t+1)(2t-5)+t(t+1) = (t+1)(2t-5+t) = (t+1)(3t-5)=0
si t+1 = 0 pour t = -1 ou 3t-5 = 0 pour t = 5/3
b) (10t+1)²-(10t+1)(t+5) = (10t+1)(10t+1-t-5) = (10t+1)(9t-4) = 0
si 10t+1 = 0 pour t = -1/10 ou 9t-4 = 0 pour t = 4/9
c) (t+3)(3t-6)+(t+3)(t-3) = (t+3)(3t-6+t-3) = (t+3)(4t-9) = 0
si t+3 = 0 pour t = -3 ou 4t-9 =0 pour t = 9/4
d) x(x+4)+(x+4)² = (x+4)(x+x+4) = (x+4)(2x+4)
si x+4 = 0 pour x = -4 ou 2x+4 = 0 pour x = -4/2 = -2
e) (7-x)(7+x)-(7+x)(2x-3) = (7+x)(7-x-2x+3) = (7+x)(10-3x) = 0
si 7+x = 0 pour x = -7 ou 10-3x = 0 pour x = -10/-3 = 10/3
f) -5y(y-3)+(y-3)² = (y-3)(-5y+y-3) = (y-3)(-4y-3) = 0
si y-3 = 0 pour y = 3 ou -4y-3 = 0 pour y = 3/-4 = -3/4
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Bonsoira) (t+1)(2t-5)+t(t+1) = (t+1)(2t-5+t) = (t+1)(3t-5)=0
si t+1 = 0 pour t = -1 ou 3t-5 = 0 pour t = 5/3
b) (10t+1)²-(10t+1)(t+5) = (10t+1)(10t+1-t-5) = (10t+1)(9t-4) = 0
si 10t+1 = 0 pour t = -1/10 ou 9t-4 = 0 pour t = 4/9
c) (t+3)(3t-6)+(t+3)(t-3) = (t+3)(3t-6+t-3) = (t+3)(4t-9) = 0
si t+3 = 0 pour t = -3 ou 4t-9 =0 pour t = 9/4
d) x(x+4)+(x+4)² = (x+4)(x+x+4) = (x+4)(2x+4)
si x+4 = 0 pour x = -4 ou 2x+4 = 0 pour x = -4/2 = -2
e) (7-x)(7+x)-(7+x)(2x-3) = (7+x)(7-x-2x+3) = (7+x)(10-3x) = 0
si 7+x = 0 pour x = -7 ou 10-3x = 0 pour x = -10/-3 = 10/3
f) -5y(y-3)+(y-3)² = (y-3)(-5y+y-3) = (y-3)(-4y-3) = 0
si y-3 = 0 pour y = 3 ou -4y-3 = 0 pour y = 3/-4 = -3/4