tische
Funktionen
Seite 111, Nr. 14, b):
| [(2x+1)/(x2-9)] - [1/(x-3)] = (2x-4)/[(x+3)2] | * [(x-3)*(x+3)*(x+3)] | ||
| [(2x+1)*(x+3)] - [(x+3)2] = (2x-4)*(x-3) | |||
| 2x2+6x+x+3-(x2+6x+9) = 2x2-6x-4x+12 | |||
| -x2+11x-18 = 0 | * (-1) | ||
| x2-11x+18 = 0 | |||
x1;2 = 11/2 ± sqr(49/4)
x1;2 = 11/2 ± 7/2
x1 = 9 und x2 = 2
Nullstellen sind x1 = 9 und x2 = 2.