URGENT s'il vous plait j'ai besoin d'aide pour mon devoir maison exercie 1 calculer à l'aide d'une identité remarquable 101x99 , 999x1001 , 37x43 , 102² -98² ,

101x99 = (100+1)(100-1) = 100²-1 = 10000-1 = 9999
999x1001 = (1000-1)(1000+1) = 1000²-1 = 1 000 000-1 = 999 999
37x43 = (40-3)(40+3) = 40²-3² = 1600-9 = 1591
102²-98² = (102+98)(102-98) = 200x4 = 800
23²-17² = (23+17)(23-17) = 40x6 = 240

ex2
1° 
(3x-1)²-16 = (3x-1+4)(3x-1-4) = (3x+3)(3x-5)=3(x+1)(3x-5)
2°
25x²-(2x+1)² = (5x+2x+1)(5x-2x+1) = (7x+1)(3x+1)
3°
(7-4x)²-9x² = (7-4x+3x)(7-4x-3x) = (7-x)(7-7x) = 7(7-x)(1-x)
4°
(3x-1)-(1-x)² = (3x-1+1-x)(3x-1-1+x) = 2x(4x-2) = 4x(2x-1)
5°
-(3+2x)²+36 = (6+3+2x)(6-3-2x) = (9+2x)(3-2x)
6°
4(2x+1)²-9(x+5)² = (2(2x+1)+3(x+5))(2(2x+1)-3(x+5)) = (4x+2+3x+15)(4x+2-3x-15)
 = (7x+17)(x-13)

Exercice 1 :

101x99 = (100+1)x(100-1)             (on applique la double distributivité)
           = 10000 - 100 + 100 - 1    (100-100=0)
           = 10000 - 1
           = 9999

999x1001 = (1000-1)x(1000+1)
               = 1000000 + 1000 - 1000 - 1
               = 1000000 - 1
               = 999999

37x43 = (40-3)x(40+3)
          = 1600 - 120 + 120 - 9
          = 1600 - 9
          = 1591

102²-98² = (100+2)² - (100-2)²
             = 10404 - 9604
             = 800

23²-17² = (20+3)² - (20-3)²
           = 529 - 289
           = 240

Exercice 2 :

1)  (3x-1)² -16 
 = (3x-1+4)(3x-1-4)
= (3x+3)(3x-5)
=3(x+1)(3x-5)

2) 25x²-(2x+1)²
= (5x+2x+1)(5x-2x+1)
= (7x+1)(3x+1)3°(7-4x)²-9x²
= (7-4x+3x)(7-4x-3x)
= (7-x)(7-7x)
= 7(7-x)(1-x)

4) (3x-1)-(1-x)²
= (3x-1+1-x)(3x-1-1+x)
= 2x(4x-2)
= 4x(2x-1)

5) -(3+2x)²+36
= (6+3+2x)(6-3-2x)
= (9+2x)(3-2x)

6) 4(2x+1)²-9(x+5)²
= (2(2x+1)+3(x+5))(2(2x+1)-3(x+5))
= (4x+2+3x+15)(4x+2-3x-15) 
= (7x+17)(x-13)