Material zum Vorkurs P2 WiSe 2013/14

WiSe 2013/14
23.09.2013
11. Tag – Exponential- und Logarithmusfunktion
Lösungen zu Polynome II
3.) a)
A
x + 2 +
B A(x − 1) + B(x + 2) (A + B)x − 1A + 2B
= =
x − 1 (x + 2)(x − 1) (x + 2)(x − 1)
!
=
−x − 5
(x + 2)(x − 1)
⇒ (A + B)x = −x ∧ −1A + 2B = −5
⇒ A + B = −1 ∧ −1A + 2B = −5
⇒ A = −1 − B ∧ 3B = −6
⇒ A = −1 − B ∧ B = −2
=⇒ A = 1 ∧ B = −2
b)
c)
=⇒ −x − 5
(x + 2)(x − 1) = 1
x + 2 + −2
x − 1 = 1
x + 2 − 2
x − 1
A
x + 1 +
B A(x + 1) + B
=
(x + 1) 2 (x + 1) 2 = Ax + A + B
(x + 1) 2 =
x
(x + 1) 2
⇒ Ax = x ∧ A + B = 0 =⇒ A = 1 ∧ B = −1
=⇒ x
(x + 1) 2 = 1
x + 1 + −1
(x + 1) 2 = 1
x + 1 − 1
(x + 1) 2
A
x − 5 +
B A(x + 5) + B(x − 5) (A + B)x + 5A − 5B
= =
x + 5 (x − 5)(x + 5)
x 2 − 25
!
= 4x − 5
x 2 − 25
⇒ (A + B)x = 4x ∧ 5A − 5B = −5
⇒ A + B = 4 ∧ A − B = −1
⇒ A + B = 4 ∧ 2A = 3
⇒ A + B = 4 ∧ A = 3 2
=⇒ A = 3 2
∧ B = 5 2
d)
=⇒ 4x − 5 3 5
x 2 − 25 = 2
x − 5 + 2
x + 5
A
x − 2 + B A(x − 2) + B Ax − 2A + B
=
(x − 2) 2 (x − 2) 2 =
(x − 2) 2 = 2x + 4
(x − 2) 2
=⇒ 2x + 4
(x − 2) 2 = 2
x − 2 + 8
(x − 2) 2
⇒ Ax = 2x ∧ −2A + B = 4
⇒ A = 2 ∧ −2A + B = 4
=⇒ A = 2 ∧ B = 8
Mathematik Vorkurs P2
✞ ☎
✝92 ✆
Dipl. Math. Stefan Podworny