Calcul littéral : développement, factorisation, identités remarquables

Savoir développer en utilisant les identités remarquables - Exercice 3

20 min
35
Développer et réduire les expressions suivantes :
Question 1

A=(7x+4)2A=\left(7x+4\right)^{2}

Correction
  • (a+b)2=a2+2ab+b2\left({\color{blue}a}+{\color{red}b}\right)^{2} ={\color{blue}a}^{2} +2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}
Ici nous avons a=7xa={\color{blue}7x} et b=4b={\color{red}4}.
A=(7x+4)2A=\left({\color{blue}7x}+{\color{red}4}\right)^{2} équivaut successivement à :
A=(7x)2+2×7x×4+42A=\left({\color{blue}7x}\right)^{2} +2\times {\color{blue}7x}\times {\color{red}4}+{\color{red}4}^{2}
A=49x2+56x+16A=49x^{2} +56x+16

Question 2

B=(4x−6)2B=\left(4x-6\right)^{2}

Correction
  • (a−b)2=a2−2ab+b2\left({\color{blue}a}-{\color{red}b}\right)^{2} ={\color{blue}a}^{2} -2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}
Ici nous avons a=4xa={\color{blue}4x} et b=6b={\color{red}6}.
B=(4x−6)2B=\left({\color{blue}4x}-{\color{red}6}\right)^{2} équivaut successivement à :
B=(4x)2−2×4x×6+62B=\left({\color{blue}4x}\right)^{2} -2\times {\color{blue}4x}\times {\color{red}6}+{\color{red}6}^{2}
B=16x2−48x+36B=16x^{2}-48x+36
Question 3

C=(5x−3)(5x+3)C=\left(5x-3\right)\left(5x+3\right)

Correction
  • (a−b)(a+b)=a2−b2\left({\color{blue}a}-{\color{red}b}\right)\left({\color{blue}a}+{\color{red}b}\right) ={\color{blue}a}^{2} -{\color{red}b}^{2}
Ici nous avons a=5xa={\color{blue}5x} et b=3b={\color{red}3}.
C=(5x−3)(5x+3)C=\left({\color{blue}5x}-{\color{red}3}\right)\left({\color{blue}5x}+{\color{red}3}\right)
C=(5x)2−(3)2C=\left({\color{blue}5x}\right)^{2} -\left({\color{red}3}\right)^{2}
C=25x2−9C=25x^{2} -9
Question 4

D=(6x+3)2D=\left(6x+3\right)^{2}

Correction
  • (a+b)2=a2+2ab+b2\left({\color{blue}a}+{\color{red}b}\right)^{2} ={\color{blue}a}^{2} +2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}
Ici nous avons a=6xa={\color{blue}6x} et b=3b={\color{red}3}.
D=(6x+3)2D=\left({\color{blue}6x}+{\color{red}3}\right)^{2} équivaut successivement à :
D=(6x)2+2×6x×3+32D=\left({\color{blue}6x}\right)^{2}+2\times {\color{blue}6x}\times {\color{red}3}+{\color{red}3}^{2}
D=36x2+36x+9D=36x^{2} +36x+9

Question 5

E=(8x−9)2E=\left(8x-9\right)^{2}

Correction
  • (a−b)2=a2−2ab+b2\left({\color{blue}a}-{\color{red}b}\right)^{2} ={\color{blue}a}^{2} -2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}
Ici nous avons a=8xa={\color{blue}8x} et b=9b={\color{red}9}.
E=(8x−9)2E=\left({\color{blue}8x}-{\color{red}9}\right)^{2} équivaut successivement à :
E=(8x)2−2×8x×9+92E=\left({\color{blue}8x}\right)^{2} -2\times {\color{blue}8x}\times {\color{red}9}+{\color{red}9}^{2}
E=64x2−144x+81E=64x^{2}-144x+81
Question 6

F=(2x−7)(2x+7)F=\left(2x-7\right)\left(2x+7\right)

Correction
  • (a−b)(a+b)=a2−b2\left({\color{blue}a}-{\color{red}b}\right)\left({\color{blue}a}+{\color{red}b}\right) ={\color{blue}a}^{2} -{\color{red}b}^{2}
Ici nous avons a=2xa={\color{blue}2x} et b=7b={\color{red}7}.
F=(2x−7)(2x+7)F=\left({\color{blue}2x}-{\color{red}7}\right)\left({\color{blue}2x}+{\color{red}7}\right)
F=(2x)2−(7)2F=\left({\color{blue}2x}\right)^{2} -\left({\color{red}7}\right)^{2}
F=4x2−49F=4x^{2} -49
Question 7

G=(−4x+3)2+3(x+1)G=\left(-4x+3\right)^{2} +3\left(x+1\right)

Correction
  • (a+b)2=a2+2ab+b2\left({\color{blue}a}+{\color{red}b}\right)^{2} ={\color{blue}a}^{2} +2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}
G=(−4x+3)2+3(x+1)G=\left({\color{blue}-4x}+{\color{red}3}\right)^{2} +3\left(x+1\right) équivaut successivement à :
G=(−4x)2+2×(−4x)×3+32+3×x+3×1G=\left({\color{blue}-4x}\right)^{2} +2\times {\color{blue}\left(-4x\right)}\times {\color{red}3}+{\color{red}3}^{2} +3\times x+3\times 1
G=16x2−24x+9+3x+3G=16x^{2} -24x+9+3x+3
G=16x2−21x+12G=16x^{2} -21x+12
Question 8

H=(5x−4)2−5(2x−6)H=\left(5x-4\right)^{2} -5\left(2x-6\right)

Correction
  • (a−b)2=a2−2ab+b2\left({\color{blue}a}-{\color{red}b}\right)^{2} ={\color{blue}a}^{2} -2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}
H=(5x−4)2−5(2x−6)H=\left({\color{blue}5x}-{\color{red}4}\right)^{2} -5\left(2x-6\right)
H=(5x)2−2×5x×4+42−(5×2x+5×(−6))H=\left({\color{blue}5x}\right)^{2} -2\times {\color{blue}5x}\times {\color{red}4}+{\color{red}4}^{2} -\left(5\times 2x+5\times \left(-6\right)\right)
H=25x2−40x+16−(10x−30)H=25x^{2} -40x+16-\left(10x-30\right) . A la prochaine étape, nous allons changer les signes à l'intérieur de la parenthèse car il y a le signe moins devant la parenthèse.
H=25x2−40x+16−10x+30H=25x^{2} -40x+16-10x+30
H=25x2−50x+46H=25x^{2} -50x+46