solvefor y,x^{2/3}+y^{2/3}=1

Lösung

löse nach

y, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}

Schritte zur Lösung

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1

Verschiebe

x^{\frac{2}{3}}

auf die rechte Seite

y^{\frac{2}{3}} = 1 - x^{\frac{2}{3}}

Potenziere beide Seiten der Gleichung mit

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

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

Löse

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

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}

Überprüfe die Lösungen:

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}

Die Lösungen sind

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}

so l v e f or y, x = 4 - y^{2}

2 x - 6 - x - 5 = x - 1

2 x + 8 = 3 x - 2

2 x + 8 = - 5

- 6 + x + 10 = 2