solvefor y,x^{2/3}+y^{2/3}=4

Lösung

löse nach

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

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}

Schritte zur Lösung

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

Verschiebe

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

auf die rechte Seite

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

Potenziere beide Seiten der Gleichung mit

3 : y^{2} = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

y^{2} = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

Löse

y^{2} = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} : y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

Überprüfe die Lösungen:

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}

Die Lösungen sind

y = 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}, y = - 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} {- 8 \leq x \leq 8}

(x) = - 1

x + 6 - 5 x + 1 + 1 = 0

x + 2 = 4 x - 2

\frac{1}{2 x} = \frac{1}{3}

6 x - 14 = x - 1