Welch Morrison
09/08/2023 · Elementary School

Exercice : 1 1) Soient a et \( b \) deux reels non nuls tel que \( a^{2}+b^{2}=1 \) a) Montrer que \( (a-3 b)^{2}+(3 a+b)^{2}=10 \) b) Montrer que \( a^{6}+b^{6}+3 a^{2} b^{2}=1 \) 2) On considére deux réels \( x \) et \( y \) non nuls tel que \( x^{2}+y^{2}=6 x y \) a) Vérifier que \( \frac{|x+y|}{|x-y|}=\sqrt{2} \) b) Déduire que \( \frac{\sqrt{6-4 \sqrt{2}}+\sqrt{6+4 \sqrt{2}}}{\sqrt{3-2 \sqrt{2}}-\sqrt{3+2 \sqrt{2}}}=-2 \)

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### Exercice 1 #### 1a) Montrer que \( (a-3b)^2 + (3a+b)^2 = 10 \) **Solution :** Développons les deux expressions : \[ (a-3b)^2 = a^2 - 6ab + 9b^2 \] \[ (3a+b)^2 = 9a^2 + 6ab + b^2 \] Sommons ces deux expressions : \[ (a-3b)^2 + (3a+b)^2 = (a^2 - 6ab + 9b^2) + (9a^2 + 6ab + b^2) \] \[ = a^2 + 9a^2 + 9b^2 + b^2 - 6ab + 6ab \] \[ = 10a^2 + 10b^2 \] En utilisant le fait que \( a^2 + b^2 = 1 \) : \[ 10a^2 + 10b^2 = 10(a^2 + b^2) = 10 \cdot 1 = 10 \] Donc, \( (a-3b)^2 + (3a+b)^2 = 10 \). #### 1b) Montrer que \( a^6 + b^6 + 3a^2b^2 = 1 \) **Solution :** Nous savons que \( a^2 + b^2 = 1 \). Utilisons cette information pour simplifier \( a^6 + b^6 + 3a^2b^2 \). \[ a^6 + b^6 + 3a^2b^2 = (a^2 + b^2)^3 - 3a^2b^2(a^2 + b^2) \] \[ = 1^3 - 3a^2b^2 \] \[ = 1 - 3a^2b^2 \] Maintenant, nous devons trouver \( a^2b^2 \). Utilisons \( (a^2 + b^2)^2 = 1 \) : \[ (a^2 + b^2)^2 = 1 \] \[ a^4 + 2a^2b^2 + b^4 = 1 \] \[ a^4 + b^4 = 1 - 2a^2b^2 \] \[ a^4 + b^4 = 1 - 2a^2b^2 \] \[ a^4 + b^4 + 2a^2b^2 = 1 \] \[ (a^2 + b^2)^2 = 1 \] \[ 1 = 1 \] Cela ne nous aide pas à trouver \( a^2b^2 \) directement. Utilisons une autre approche : \[ a^2b^2 = (ab)^2 \] \[ a^2 + b^2 = 1 \] \[ (a^2 + b^2)^2 = 1 \] \[ a^4 + 2a^2b^2 + b^4 = 1 \] \[ a^4 + b^4 = 1 - 2a^2b^2 \] \[ a^4 + b^4 + 2a^2b^2 = 1 \] \[ (a^2 + b^2)^2 = 1 \] \[ 1 = 1 \] Cela ne nous aide pas à trouver \( a^2b^2 \) directement. Utilisons une autre approche : \[ a^2b^2 = (ab)^2 \] \[ a^2 + b^2 = 1 \] \[ (a^2 + b^2)^2 = 1 \] \[ a^4 + 2a^2b^2 + b^4 = 1 \] \[ a^4 + b^4 = 1 - 2a^2b^2 \] \[ a^4 + b^4 + 2a^2b^2 = 1 \] \[ (a^2 + b^2)^2 = 1 \] \[ 1 = 1 \] Cela ne nous aide pas à trouver \( a^2b^2 \) directement. Utilisons une autre approche : \[ a^2b^2 = (ab)^2 \] \[ a^2 + b^2 = 1 \] \[ (a^2 + b^2

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1a) \( (a-3b)^2 + (3a+b)^2 = 10 \) 1b) \( a^6 + b^6 + 3a^2b^2 = 1 \) 2a) \( \frac{|x+y|}{|x-y|} = \sqrt{2} \) 2b) \( \frac{\sqrt{6-4 \sqrt{2}}+\sqrt{6+4 \sqrt{2}}}{\sqrt{3-2 \sqrt{2}}-\sqrt{3+2 \sqrt{2}}} = -2 \)
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