Chapitre 2 Calcul littéral

a b k
k×(a+b) = ka+kb
15 ×12 = 15 ×(10 + 2) 7(2x+ 5) = 7×2x+7×5
=15 ×10+15 ×2 = 14x+ 35
= 150 + 30 = 180
AABCD =AB ×BC = (a+b)×(c+d)
AABCD =AAEI G +AEBH I +AIH CF +AGI F D ,
AABCD =ac +bc +bd +ad
a b c d
(a+b)×(c+d) = ac+ad+bc+bd

32 ×24 = (30 +2)(20 + 4)
=30 ×20+30 ×4+2×20+2×4
= 600 + 120 + 40 + 8
= 768
(x+2)(2x−3) = x×2x−x×3+2×2x−2×3
= 2x2−3x+ 4x−6
= 2x2+x−6
ab
(a+b)2=a2+ 2 ×a×b+b2
(x+3)2=x2+ 2 ×x×3+32=x2+ 6x+ 9
(5x+4)2= (5x)2+ 2 ×5x×4+42= 25x2+ 40x+ 16
3
2x+52=3 2x2+ 2 ×3 2x×5+52=9 4x2+ 15x+ 25
(a+b)2=a2+ 2 ×a×b+b2(a+b)2=a2+ 2ab+b2
a2b22ab
2ab
ab
(a−b)2=a2−2×a×b+b2
(x−3)2=x2−2×x×3+32=x2−6x+ 9
(5x−4)2= (5x)2−2×5x×4+42= 25x2−40x+ 16
3
2x−52=3 2x2−2×3 2x×5+52=9 4x2−15x+ 25

ab
(a−b)(a+b) = a2−b2
(x−3)(x+3) = x2−32=x2−9
(5x+4)(5x−4) = (5x)2−42= 25x2−16
3
2x−53 2x+5=3 2x2−52=9 4x2−25
A2−B2
a2+ 2ab +b2a2−2ab +b2
4x2+ 12x+ 9
a2+ 2ab+b2
4x2+ 12x+ 9 = (2x)2+ 2 ×2x×3+322x a 3b
4x2+ 12x+ 9 = (2x+3)2
25 −(x+ 1)2
25 −(x+ 1)2=52−(x+ 1)2
A2−B25Ax+ 1 B

52−(x+ 1)2= (5−(x+ 1)) (5+(x+ 1))
= (5 −x−1)(5 + x+ 1)
= (4 −x)(6 + x)
(5x−2)(x+ 3) −(x+ 3)(2x−3)
(x+ 3)
(5x−2)(x+ 3) −(x+ 3)(2x−3) = (x+ 3) ((5x−2) −(2x−3))
=(x+ 3) (5x−2−2x+ 3)
=(x+ 3)(3x+ 1)
4x+ 4 + (x+ 1)2
a2+ 2ab +b2
4x+ 4 + (x+ 1)2=4×x+4×1+(x+ 1)2=4×(x+ 1) + (x+ 1) ×(x+ 1)
(x+ 1)
4x+ 4 + (x+ 1)2= 4 ×(x+1)+ (x+ 1) ×(x+ 1)
=(x+ 1) (4 + (x+ 1))
=(x+ 1)(x+ 5)