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Lösungsbeispiel Bruchgleichung 7:

3 2x. 3 x. 9 8. +. =. 3 2x. 3 x. 9 8. +. =. • 8x. 3 • 8x 2x. 3 • 8x x. 9 • 8x 8. +. =. T. 12 + 24 = 9x. T. 36 = 9x. : 9. Lösungsbeispiel Bruchgleichung 7:. Die Lösung x = 0 müssen wir ausschließen also ist die Definitionsmenge D = Q {0}.

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Lösungsbeispiel Bruchgleichung 7:

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  1. 3 2x 3 x 9 8 + = 3 2x 3 x 9 8 + = • 8x 3 • 8x 2x 3 • 8x x 9 • 8x 8 + = T 12 + 24 = 9x T 36 = 9x : 9 Lösungsbeispiel Bruchgleichung 7: • Die Lösung x = 0 müssen wir ausschließen also ist • die Definitionsmenge D = Q \ {0} 2. Bestimmung des Hauptnenners: 2x = 2 • x x = x 8 = 2 • 2 • 2 HN = 2 • 2 • 2 • x Lösungsschritte: 4 = x L = { 4 }

  2. x 2x+4 1 x+2 5 4x+8 + = 1 x+2 x 2(x+2) 5 4(x+2) + = • 4(x+2) x • 4(x+2) 2(x+2) 1 • 4(x+2) x+2 5 • 4(x+2) 4(x+2) T = + - 4 2x + 4 = 5 2x = 1 : 2 Lösungsbeispiel Bruchgleichung 8: • Die Lösung x = -2 müssen wir ausschließen also ist • die Definitionsmenge D = Q \ { -2 } 2. Bestimmung des Hauptnenners: 2x+4 = 2 • (x+2) In diesem Fall erkennt man den x+2 = x+2 Hauptnenner besser, wenn man die 4x+8 = 2 • 2 (x+2) einzelnen Nenner faktorisiert ! HN = 2 • 2 • (x+2) Lösungsschritte: x = 0,5 L = { 0,5 }

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