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Cocotebleu
@Cocotebleu
April 2019
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Montrer que pour tout réel x différent de 6 et de 7 : x-5/x-6 - x-6/x-7 = -1/(x-6)(x-7) sachant que a/b+c/d = ad+bc/bd
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kisimoha
Bonsoir,
(x-5)/(x-6) - (x-6)/(x-7) =
[
(x-5)(x-7) -(x-6)(x-6)
]/
(x-6)(x-7)
=
[
(x² -5x -7x +35) -(x² -6x-6x +36)
]/
(x-6)(x-7)
=( x² -12x +35 -x² +12x -36)
/
(x-6)(x-7)
=(-1)
/
(x-6)(x-7) ( on a x²-x²=0 et -12x+12x=0)
donc (x-5)
/
(x-6) - (x-6)
/
(x-7) = -1
/
(x-6)(x-7).
0 votes
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(x-5)/(x-6) - (x-6)/(x-7) = [(x-5)(x-7) -(x-6)(x-6)]/(x-6)(x-7)
= [(x² -5x -7x +35) -(x² -6x-6x +36)]/(x-6)(x-7)
=( x² -12x +35 -x² +12x -36)/(x-6)(x-7)
=(-1)/(x-6)(x-7) ( on a x²-x²=0 et -12x+12x=0)
donc (x-5)/(x-6) - (x-6)/(x-7) = -1/(x-6)(x-7).