solvefor y,z=yx^6-3x+8x^2y^2-y^2

Lösung

löse nach

y, z = y x^{6} - 3 x + 8 x^{2} y^{2} - y^{2}

y = \frac{- x ^{6} + x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}, y = \frac{- x ^{6} - x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}; x \neq = \frac{1}{2 2}, x \neq = - \frac{1}{2 2}

Schritte zur Lösung

z = y x^{6} - 3 x + 8 x^{2} y^{2} - y^{2}

Tausche die Seiten

y x^{6} - 3 x + 8 x^{2} y^{2} - y^{2} = z

Verschiebe

z

auf die linke Seite

y x^{6} - 3 x + 8 x^{2} y^{2} - y^{2} - z = 0

Schreibe in der Standard Form

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

(8 x^{2} - 1) y^{2} + x^{6} y - 3 x - z = 0

Löse mit der quadratischen Formel

y_{1, 2} = \frac{- x ^{6} \pm ( x ^{6} ) ^{2} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}

Vereinfache

(x^{6})^{2} - 4 (8 x^{2} - 1) (- 3 x - z) : x^{12} - 4 (8 x^{2} - 1) (- 3 x - z)

y_{1, 2} = \frac{- x ^{6} \pm x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}; x \neq = \frac{1}{2 2}, x \neq = - \frac{1}{2 2}

Trenne die Lösungen

y_{1} = \frac{- x ^{6} + x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}, y_{2} = \frac{- x ^{6} - x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}

y = \frac{- x ^{6} + x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}; x \neq = \frac{1}{2 2}, x \neq = - \frac{1}{2 2}

y = \frac{- x ^{6} - x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}; x \neq = \frac{1}{2 2}, x \neq = - \frac{1}{2 2}

Die Lösungen für die quadratische Gleichung sind:

y = \frac{- x ^{6} + x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}, y = \frac{- x ^{6} - x ^{12} - 4 ( 8 x ^{2} - 1 ) ( - 3 x - z )}{2 ( 8 x ^{2} - 1 )}; x \neq = \frac{1}{2 2}, x \neq = - \frac{1}{2 2}

2 x^{2} - 2 = - 3 x

6 x + 3 = 2 x^{2}

co m pl e t es q u a re x^{2} - \frac{4}{5} x - 1

co m pl e t es q u a re z^{2} - 2 z

z^{2} + 3 z - 7 = z