solvefor y,x^3+x^2y^2-4x-2y=0

Lösung

löse nach

y, x^{3} + x^{2} y^{2} - 4 x - 2 y = 0

y = \frac{1 + - x ^{3} ( x ^{2} - 4 ) + 1}{x ^{2}}, y = \frac{1 - - x ^{3} ( x ^{2} - 4 ) + 1}{x ^{2}}; x \neq = 0

Schritte zur Lösung

x^{3} + x^{2} y^{2} - 4 x - 2 y = 0

Schreibe in der Standard Form

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

x^{2} y^{2} - 2 y + x^{3} - 4 x = 0

Löse mit der quadratischen Formel

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

Vereinfache

(- 2)^{2} - 4 x^{2} (x^{3} - 4 x) : 2 1 - x^{3} (x^{2} - 4)

y_{1, 2} = \frac{- ( - 2 ) \pm 2 1 - x ^{3} ( x ^{2} - 4 )}{2 x ^{2}}; x \neq = 0

Trenne die Lösungen

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

y = \frac{- ( - 2 ) + 2 1 - x ^{3} ( x ^{2} - 4 )}{2 x ^{2}} : \frac{1 + - x ^{3} ( x ^{2} - 4 ) + 1}{x ^{2}}

y = \frac{- ( - 2 ) - 2 1 - x ^{3} ( x ^{2} - 4 )}{2 x ^{2}} : \frac{1 - - x ^{3} ( x ^{2} - 4 ) + 1}{x ^{2}}

Die Lösungen für die quadratische Gleichung sind:

y = \frac{1 + - x ^{3} ( x ^{2} - 4 ) + 1}{x ^{2}}, y = \frac{1 - - x ^{3} ( x ^{2} - 4 ) + 1}{x ^{2}}; x \neq = 0

2 x^{2} + 4 x - 39 = x^{2}

x^{2} - 14 x + 4 = - 10 x + 9

x^{2} - x + 1 = 2

\frac{1}{6} n^{2} - \frac{1}{2} n - \frac{2}{3} = 0

so l v e f or a, (\frac{a}{2} + r)^{2} = (\frac{a}{2} - r)^{2} + (\frac{a}{2} - r)^{2}