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Mademoiselle3
@Mademoiselle3
April 2019
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K=(5x-2)*-(x-7)(5x-2)
1)Développer et réduire k
2)Factoriser k
3)En choisissant la forme de k la plus adaptée parmi celle trouvées, montrer que 2/5 est une valeur qui annulé k
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Omnes
Salut,
k=(5x-2)*-(x-7)(5x-2)
1)Développer et réduire k
k = (5x)² - 2*2*5x + 2²- [5x² - 2x - 35x + 14]
k = 25x² - 20x + 4 -5x² +37x - 14
k = 20x² + 17x - 10
2)Factoriser k
k=(5x-2)*-(x-7)(5x-2)
k=(5x-2)(5x-2) - (x-7)(5x-2)
k= (5x-2)[(5x-2) - (x-7)]
k=(5x-2)(5x-2-x+7)
k = (5x-2)(4x+5)
3)En choisissant la forme de k la plus adaptée parmi celle trouvées, montrer que 2/5 est une valeur qui annulé k
k = (5x-2)(4x+5)
On remplace x par 2/5
k = (5*2/5 - 2)(4 * 2/5 + 5)
k = (2-2)(8/5 + 5)
k = 0*33/5
k = 0.
Bonne soirée !
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k=(5x-2)*-(x-7)(5x-2)
1)Développer et réduire k
k = (5x)² - 2*2*5x + 2²- [5x² - 2x - 35x + 14]
k = 25x² - 20x + 4 -5x² +37x - 14
k = 20x² + 17x - 10
2)Factoriser k
k=(5x-2)*-(x-7)(5x-2)
k=(5x-2)(5x-2) - (x-7)(5x-2)
k= (5x-2)[(5x-2) - (x-7)]
k=(5x-2)(5x-2-x+7)
k = (5x-2)(4x+5)
3)En choisissant la forme de k la plus adaptée parmi celle trouvées, montrer que 2/5 est une valeur qui annulé k
k = (5x-2)(4x+5)
On remplace x par 2/5
k = (5*2/5 - 2)(4 * 2/5 + 5)
k = (2-2)(8/5 + 5)
k = 0*33/5
k = 0.
Bonne soirée !