Sagot :
1)
B = (2x-3)(5x+7)-2x+3 = (2x-3)(5x+7)-1(2x-3)
B = (2x-3)(5x+7-1)=(2x-3)(5x+6)
2)
C = (x-3)² - x + 3 = (x-3)² - 1(x-3) = (x-3)(x-3-1) = (x-3)(x-4)C = (x-3)² - x + 3 = (x-3)² - 1(x-3) = (x-3)(x-3-1) = (x-3)(x-4)
B = (2x-3)(5x+7)-2x+3 = (2x-3)(5x+7)-1(2x-3)
B = (2x-3)(5x+7-1)=(2x-3)(5x+6)
2)
C = (x-3)² - x + 3 = (x-3)² - 1(x-3) = (x-3)(x-3-1) = (x-3)(x-4)C = (x-3)² - x + 3 = (x-3)² - 1(x-3) = (x-3)(x-3-1) = (x-3)(x-4)
B=(2x-3)(5x+7)-2x+3 B=(2x-3)(5x+7)-1....
(-2x+3)=-1(2x-3)
Donc B=(2x-3)(5x+7)-1(2x-3)
b)
(2x-3)[(5x+7)-1]
2.Factoriser :C= (x-3)²-x+3
C= (x-3)²+(x-3)
C=(x-3)[(x-3)+1]
D = ( x-4)(3x+2)-3x-2
D= (x-4)(3x+2)+(3x+2)
D= (3x-2)[(x-4)+1]
J'espere ne pas avoir fait de betises
(-2x+3)=-1(2x-3)
Donc B=(2x-3)(5x+7)-1(2x-3)
b)
(2x-3)[(5x+7)-1]
2.Factoriser :C= (x-3)²-x+3
C= (x-3)²+(x-3)
C=(x-3)[(x-3)+1]
D = ( x-4)(3x+2)-3x-2
D= (x-4)(3x+2)+(3x+2)
D= (3x-2)[(x-4)+1]
J'espere ne pas avoir fait de betises