solvefor y,x^2y-xy^2+x^2+y^2=0

Lösung

löse nach

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

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

Schritte zur Lösung

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

Schreibe in der Standard Form

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

(- x + 1) y^{2} + x^{2} y + x^{2} = 0

Löse mit der quadratischen Formel

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

Vereinfache

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

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

Trenne die Lösungen

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

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

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

Die Lösungen für die quadratische Gleichung sind:

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

2 (x + 1)^{2} = 12

x^{2} = 32 x + 33

x^{2} + 4 0^{2} = 4 1^{2}

- x^{2} + 6 x + 16 = 0

so l v e f or s, s^{2} + 2 s + 2 = 0