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Younes95
@Younes95
May 2019
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Bonjour j'ai besoin d'aide pour c'est calcule svp
Factoriser (et reduire)
a. 64 + 112 y+ 49y²
b. 4x² - 25
c. (x +4)(2x +9) + (x+4)(3x - 5)
d. (2 - y)(4y + 10) - (4y + 10)(-8 -3y)
e. (3y -7)(2y+ 8)+(3y-7)(5y-11)
f. x² + 10x + 25
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aymanemaysae
Verified answer
Bonjour ;
a)
64 + 112 y+ 49y² = 8² + 2 x 8 x 7y + (7y)² = (8 + 7y)² .
b)
4x² - 25 = (2x)² - 5² = (2x - 5)(2x + 5) .
c)
(x +4)(2x +9) + (x+4)(3x - 5) = (x + 4)(2x + 9 + 3x -5) = = (x + 4)(5x + 4) .
d)
(2 - y)(4y + 10) - (4y + 10)(-8 -3y) = (4y + 10)(2 - y + 8 + 3y)
= (4y + 10)(2y + 10) = 4(2y + 5)(y + 5) .
e)
(3y -7)(2y+ 8)+(3y-7)(5y-11) = (3y - 7)(2y + 8 + 5y - 11) = (3y - 7)(7y - 3) .
f)
x² + 10x + 25 = x² + 2 * 5 * x + 5² = (x + 5)² .
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gsantha
Bonjour voilà la réponse :
a. 64 + 112 y+ 49y² = (7y+8)²
car (7y)²+2*7y*8+8² = 49y²+112y+64 => identité remarquable (a+b)² = a²+2ab+b²
b.
4x² - 25 = (2x+5)(2x-5)
car 2x*2x+2x*(-5)+5*2x-5*5 = 4x²-10x+10x-25 = 4x²-25 => identité remarquable (a+b)(a-b) = a²-b²
c.
(x +4)(2x +9) + (x+4)(3x - 5) = (x+4)[(2x+9)+(3x-5)]
= (x+4)(2x+9+3x-5) = (x+4)(5x+4) => facteur commun
d.
(2 - y)(4y + 10) - (4y + 10)(-8 -3y) = (4y+10)[(2-y)-(-8-3y)]
= (4y+10)(2-y-8+3y) = (4y+10)(2y-6) => facteur commun
e.
x² + 10x + 25 = (x+5)²
car (x+5)² = x²+2*x*5+5² = x²+10x+25 => identité remarquable (a+b)² = a²+2ab+b²
Voilà j'espère avoir pu t'aider
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Verified answer
Bonjour ;a)
64 + 112 y+ 49y² = 8² + 2 x 8 x 7y + (7y)² = (8 + 7y)² .
b)
4x² - 25 = (2x)² - 5² = (2x - 5)(2x + 5) .
c)
(x +4)(2x +9) + (x+4)(3x - 5) = (x + 4)(2x + 9 + 3x -5) = = (x + 4)(5x + 4) .
d)
(2 - y)(4y + 10) - (4y + 10)(-8 -3y) = (4y + 10)(2 - y + 8 + 3y)
= (4y + 10)(2y + 10) = 4(2y + 5)(y + 5) .
e)
(3y -7)(2y+ 8)+(3y-7)(5y-11) = (3y - 7)(2y + 8 + 5y - 11) = (3y - 7)(7y - 3) .
f)
x² + 10x + 25 = x² + 2 * 5 * x + 5² = (x + 5)² .
a. 64 + 112 y+ 49y² = (7y+8)²
car (7y)²+2*7y*8+8² = 49y²+112y+64 => identité remarquable (a+b)² = a²+2ab+b²
b. 4x² - 25 = (2x+5)(2x-5)
car 2x*2x+2x*(-5)+5*2x-5*5 = 4x²-10x+10x-25 = 4x²-25 => identité remarquable (a+b)(a-b) = a²-b²
c. (x +4)(2x +9) + (x+4)(3x - 5) = (x+4)[(2x+9)+(3x-5)]
= (x+4)(2x+9+3x-5) = (x+4)(5x+4) => facteur commun
d. (2 - y)(4y + 10) - (4y + 10)(-8 -3y) = (4y+10)[(2-y)-(-8-3y)]
= (4y+10)(2-y-8+3y) = (4y+10)(2y-6) => facteur commun
e. x² + 10x + 25 = (x+5)²
car (x+5)² = x²+2*x*5+5² = x²+10x+25 => identité remarquable (a+b)² = a²+2ab+b²
Voilà j'espère avoir pu t'aider