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Eden192
@Eden192
April 2019
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Bonjour !
pourriez vous m'aider à résoudre cette operation ? (je crois que j'ai mis beaucoup de parenthese inutile..)
[(2^(n+3))/(3^(n+1))]*[3^n/2^(n+2)]
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[(2^(n+3))/(3^(n+1))]*[3^n/2^(n+2)]
=2^(n+3-n-2)*3^(n-n-1)
=2^1*3^(-1)
=2/3
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=2^(n+3-n-2)*3^(n-n-1)
=2^1*3^(-1)
=2/3