Ch11 22
- 1. Exercice 22
λ −1 x 2λ
· =
1 λ y 1 − λ2
−1
x λ −1 2λ
= ·
y 1 −λ 1 − λ2
Comme : −1
λ
1
λ −1 λ2 +1 λ2 +1
= −1 λ
1 −λ λ2 +1 λ2 +1
Alors :
- 2. Exercice 22 (suite..)
λ
1
x λ2 +1 λ2 +1
2λ
= −1 λ
·
1 − λ2
y λ2 +1 λ2 +1
λ +1
2
1−λ2
2
2λ
λ2 +1 λ2 +1 λ2 +1
= =
−2λ λ−λ3 −λ−λ3
λ2 +1 λ2 +1 λ2 +1
1
1
= −λ(λ2 +1) =
λ2 +1
−λ
Donc : x = 1 et y = −λ