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
Quick Answer
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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