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Beliebt Algebra >

solvefor a,(a+b+c)(1+ab+ac+bc)=32

Lösung

löse nach

a, (a + b + c) (1 + ab + a c + b c) = 32

Lösung

a = \frac{- b ^{2} - 3 b c - 1 - c ^{2} + b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}, a = \frac{- b ^{2} - 3 b c - 1 - c ^{2} - b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}; b + c \neq = 0

Schritte zur Lösung

(a + b + c) (1 + ab + a c + b c) = 32

Schreibe

(a + b + c) (1 + ab + a c + b c)

um:

a^{2} b + a b^{2} + 3 ab c + b^{2} c + b c^{2} + b + a + a^{2} c + c + a c^{2}

a^{2} b + a b^{2} + 3 ab c + b^{2} c + b c^{2} + b + a + a^{2} c + c + a c^{2} = 32

Verschiebe

32

auf die linke Seite

a^{2} b + a b^{2} + 3 ab c + b^{2} c + b c^{2} + b + a + a^{2} c + c + a c^{2} - 32 = 0

Schreibe in der Standard Form

a x^{2} + b x + c = 0

(b + c) a^{2} + (b^{2} + 3 b c + 1 + c^{2}) a + b^{2} c + b c^{2} + b + c - 32 = 0

Löse mit der quadratischen Formel

a_{1, 2} = \frac{- ( b ^{2} + 3 b c + 1 + c ^{2} ) \pm ( b ^{2} + 3 b c + 1 + c ^{2} ) ^{2} - 4 ( b + c ) ( b ^{2} c + b c ^{2} + b + c - 32 )}{2 ( b + c )}

Vereinfache

(b^{2} + 3 b c + 1 + c^{2})^{2} - 4 (b + c) (b^{2} c + b c^{2} + b + c - 32) : b^{4} + 2 b^{3} c + 3 b^{2} c^{2} - 2 b^{2} + 2 b c^{3} - 2 b c + 128 b + c^{4} - 2 c^{2} + 128 c + 1

a_{1, 2} = \frac{- ( b ^{2} + 3 b c + 1 + c ^{2} ) \pm b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}; b + c \neq = 0

Trenne die Lösungen

a_{1} = \frac{- ( b ^{2} + 3 b c + 1 + c ^{2} ) + b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}, a_{2} = \frac{- ( b ^{2} + 3 b c + 1 + c ^{2} ) - b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}

a = \frac{- ( b ^{2} + 3 b c + 1 + c ^{2} ) + b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )} : \frac{- b ^{2} - 3 b c - 1 - c ^{2} + b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}

a = \frac{- ( b ^{2} + 3 b c + 1 + c ^{2} ) - b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )} : \frac{- b ^{2} - 3 b c - 1 - c ^{2} - b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}

Die Lösungen für die quadratische Gleichung sind:

a = \frac{- b ^{2} - 3 b c - 1 - c ^{2} + b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}, a = \frac{- b ^{2} - 3 b c - 1 - c ^{2} - b ^{4} + 2 b ^{3} c + 3 b ^{2} c ^{2} - 2 b ^{2} + 2 b c ^{3} - 2 b c + 128 b + c ^{4} - 2 c ^{2} + 128 c + 1}{2 ( b + c )}; b + c \neq = 0

Graph

Beliebte Beispiele

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