Calcul littéral : développement, factorisation, identités remarquables

Savoir développer en utilisant les identités remarquables - Exercice 1

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

A=(x+7)2A=\left(x+7\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=xa={\color{blue}x} et b=7b={\color{red}7}.
A=(x+7)2A=\left({\color{blue}x}+{\color{red}7}\right)^{2} équivaut successivement à :
A=x2+2×x×7+72A={\color{blue}x}^{2} +2\times {\color{blue}x}\times {\color{red}7}+{\color{red}7}^{2}
A=x2+14x+49A=x^{2} +14x+49

Question 2

B=(2x+3)2B=\left(2x+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=2xa={\color{blue}2x} et b=3b={\color{red}3}.
B=(2x+3)2B=\left({\color{blue}2x}+{\color{red}3}\right)^{2} équivaut successivement à :
B=(2x)2+2×2x×3+32B=\left({\color{blue}2x}\right)^{2} +2\times {\color{blue}2x}\times {\color{red}3}+{\color{red}3}^{2}
B=4x2+12x+9B=4x^{2} +12x+9
Question 3

C=(3x5)2C=\left(3x-5\right)^{2}

Correction
  • (ab)2=a22ab+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=3xa={\color{blue}3x} et b=5b={\color{red}5}.
C=(3x5)2C=\left({\color{blue}3x}-{\color{red}5}\right)^{2} équivaut successivement à :
C=(3x)22×3x×5+52C=\left({\color{blue}3x}\right)^{2} -2\times {\color{blue}3x}\times {\color{red}5}+{\color{red}5}^{2}
C=9x230x+25C=9x^{2} -30x+25
Question 4

D=(6x4)2D=\left(6x-4\right)^{2}

Correction
  • (ab)2=a22ab+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=4b={\color{red}4}.
D=(6x4)2D=\left({\color{blue}6x}-{\color{red}4}\right)^{2} équivaut successivement à :
D=(6x)22×6x×4+42D=\left({\color{blue}6x}\right)^{2} -2\times {\color{blue}6x}\times {\color{red}4}+{\color{red}4}^{2}
D=36x248x+16D=36x^{2} -48x+16
Question 5

E=(3x2)(3x+2)E=\left(3x-2\right)\left(3x+2\right)

Correction
  • (ab)(a+b)=a2b2\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=3xa={\color{blue}3x} et b=2b={\color{red}2}.
E=(3x2)(3x+2)E=\left({\color{blue}3x}-{\color{red}2}\right)\left({\color{blue}3x}+{\color{red}2}\right)
E=(3x)2(2)2E=\left({\color{blue}3x}\right)^{2} -\left({\color{red}2}\right)^{2}
E=9x24E=9x^{2} -4
Question 6

F=(x1)2F=\left(x-1\right)^{2}

Correction
  • (ab)2=a22ab+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=xa={\color{blue}x} et b=1b={\color{red}1}.
F=(x1)2F=\left({\color{blue}x}-{\color{red}1}\right)^{2} équivaut successivement à :
F=x22×x×1+12F={\color{blue}x}^{2} -2\times {\color{blue}x}\times {\color{red}1}+{\color{red}1}^{2}
F=x22x+1F=x^{2} -2x+1
Question 7

G=(x+2)2+2(3x5)G=\left(x+2\right)^{2} +2\left(3x-5\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=(x+2)2+2(3x5)G=\left({\color{blue}x}+{\color{red}2}\right)^{2} +2\left(3x-5\right) équivaut successivement à :
G=x2+2×x×2+22+2×3x+2×(5)G={\color{blue}x}^{2} +2\times {\color{blue}x}\times {\color{red}2}+{\color{red}2}^{2} +2\times 3x+2\times \left(-5\right)
G=x2+4x+4+6x10G=x^{2} +4x+4+6x-10
G=x2+10x6G=x^{2} +10x-6
Question 8

H=(2x3)23(4x5)H=\left(2x-3\right)^{2} -3\left(4x-5\right)

Correction
  • (ab)2=a22ab+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=(2x3)23(4x5)H=\left({\color{blue}2x}-{\color{red}3}\right)^{2} -3\left(4x-5\right)
H=(2x)22×2x×3+32(3×4x+3×(5))H=\left({\color{blue}2x}\right)^{2} -2\times {\color{blue}2x}\times {\color{red}3}+{\color{red}3}^{2} -\left(3\times 4x+3\times \left(-5\right)\right)
H=4x212x+9(12x15)H=4x^{2} -12x+9-\left(12x-15\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=4x212x+912x+15H=4x^{2} -12x+9-12x+15
H=4x224x+24H=4x^{2} -24x+24