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quatre
@quatre
January 2021
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Bonjour mesdames et messieurs voici mon problème
Développer et réduire:
1) ( 3× - 1 ) (× - 1 )
2) 5( × - 1 ) (4 + × )
3) ( × - 1 ) ( × - 2 ) ( × - 3 )
Merci bien
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PAU64
1) (3x - 1) (x - 1)
= 3x * x + 3x * (- 1) - 1 * x - 1 * (- 1)
= 3x² - 3x - x + 1
= 3x² - 4x + 1
2) 5 (x - 1) (4 + x)
= 5 (x * 4 + x * x - 1 * 4 - 1 * x)
= 5 (4x + x² - 4 - x)
= 5 * 4x + 5 * x² + 5 * (- 4) + 5 * (- x)
= 20x + 5x² - 20 - 5x
= 5x² + 15x - 20
3) (x - 1) (x - 2) (x - 3)
= [x * x + x * (- 2) - 1 * x - 1 * (- 2)] (x - 3)
= (x² - 2x - x + 2) (x - 3)
= x * x² + x * (- 2x) + x * (- x) + x * 2 - 3 * x² - 3 * (- 2x) - 3 * (- x) - 3 * 2
= x³ - 2x² - x² + 2x - 3x² + 6x + 3x - 6
= x³ - 6x² + 11x - 6
Les * désignent "fois".
3 votes
Thanks 1
quatre
MERCIIII !!!
PAU64
de rien ;)
quatre
je ne sais pas comment vous faites mais merci encore une fois !!
PAU64
j'en ai tellement fait au collège qu'à force je maîtrise ;)
quatre
haha ! C'est sur :) merci 1000 fois
PAU64
;) de rien
quatre
;)
mjalalrahmani
1) 3X²-3X-X+1=3X²-4X+1
2) 20X+5X²-20-5X=5X²+15X-20
3) (X²-2X-X+2) (X-3) = X^3-3X²-2X²+6X-X²+3X+2X-6 = X^3-6X²+11X-6
0 votes
Thanks 1
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= 3x * x + 3x * (- 1) - 1 * x - 1 * (- 1)
= 3x² - 3x - x + 1
= 3x² - 4x + 1
2) 5 (x - 1) (4 + x)
= 5 (x * 4 + x * x - 1 * 4 - 1 * x)
= 5 (4x + x² - 4 - x)
= 5 * 4x + 5 * x² + 5 * (- 4) + 5 * (- x)
= 20x + 5x² - 20 - 5x
= 5x² + 15x - 20
3) (x - 1) (x - 2) (x - 3)
= [x * x + x * (- 2) - 1 * x - 1 * (- 2)] (x - 3)
= (x² - 2x - x + 2) (x - 3)
= x * x² + x * (- 2x) + x * (- x) + x * 2 - 3 * x² - 3 * (- 2x) - 3 * (- x) - 3 * 2
= x³ - 2x² - x² + 2x - 3x² + 6x + 3x - 6
= x³ - 6x² + 11x - 6
Les * désignent "fois".
2) 20X+5X²-20-5X=5X²+15X-20
3) (X²-2X-X+2) (X-3) = X^3-3X²-2X²+6X-X²+3X+2X-6 = X^3-6X²+11X-6