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ZentoGaming
@ZentoGaming
May 2019
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Bonjour, je suis en 1èreS je dois résoudre cette équation mais je n'y arrive pas
3B/B+2 - B+1/B-2 = -11/5
QUAND JE MET "/" ÇA SIGNIFIE LA BARRE DE FRACTION
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laurance
3b/(b+2)= (b+1)/(b-2) - 11/5 donc
15b/(b+2)= (5b+5)/(b-2) - 11(b-2)/(b-2)
=(5b+5-11b+22)/(b-2)= (-6b+27)/(b-2) on en déduit
15b(b-2)=(b+2)(-6b+27)
puis 5b(b-2)= (b+2)(-2b+9) et
5b²-10b= -2b² +5b +18
5b² - 15b = -2b² + 18 d'où
5b(b-3)= -2(b²-9) = -2(b-3)(b+3) (b-3) ( 5b + 2(b+3) )= 0
(b-3)( 7b + 6)= 0 b-3=0 ou 7b+6=0
b=3 ou b= -6/7
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15b/(b+2)= (5b+5)/(b-2) - 11(b-2)/(b-2)
=(5b+5-11b+22)/(b-2)= (-6b+27)/(b-2) on en déduit
15b(b-2)=(b+2)(-6b+27)
puis 5b(b-2)= (b+2)(-2b+9) et
5b²-10b= -2b² +5b +18
5b² - 15b = -2b² + 18 d'où
5b(b-3)= -2(b²-9) = -2(b-3)(b+3) (b-3) ( 5b + 2(b+3) )= 0
(b-3)( 7b + 6)= 0 b-3=0 ou 7b+6=0
b=3 ou b= -6/7