Aufgabe:
\( \frac{1}{|x-3|} \) + \( \frac{1}{x+3} \) = 6
Ist meine Lösung/Ansatz so richtig? :)
Problem/Ansatz:
1. Fall: x-3 ≥ 0 => x≥3
\( \frac{1}{x-3} \) + \( \frac{1}{x+3} \) = 6 | *(x-3)(x+3)
x+3 + x-3 = 6(x-3)(x+3)
2x = 6(x2-9)
2x = 6x2 - 54
6x2 -2x -54 = 0
x2 - \( \frac{1}{3} \)x - 9 = 0
x1/2 = 1/6+- \( \sqrt{1/36 + 9} \)
x1=3,17 ; x2= -2,84
L1 = {x∈ℝ | x = 3,17}
2. Fall x-3<0 x<3
\( \frac{1}{-(x-3)} \) + \( \frac{1}{x+3} \) = 6
\( \frac{1}{-x+3)} \) + \( \frac{1}{x+3} \) = 6 | *(-x+3)(x+3)
-x+3 + x+3 = 6(-x+3)(x+3)
6 = 6(-x2 -3x +3x +9)
6 = 6(-x2 +9)
6 = -6x2 +54
-6x2 + 48 = 0
-6x2 = -48
x2 = 8
x1 = \( \sqrt{8} \) ; x2 = -\( \sqrt{8} \)
L2 = {x∈ℝ| -2,83; 2,83}
Lges = {x∈ℝ| -2,83; 2,83; 3,17}