solvefor x,x^2p^2+y^2q^2=z^2

Lösung

löse nach

x, x^{2} \cdot p^{2} + y^{2} \cdot q^{2} = z^{2}

x = \frac{z ^{2} - y ^{2} q ^{2}}{p}, x = - \frac{z ^{2} - y ^{2} q ^{2}}{p}; p \neq = 0

Schritte zur Lösung

x^{2} p^{2} + y^{2} q^{2} = z^{2}

Verschiebe

y^{2} \cdot q^{2}

auf die rechte Seite

x^{2} p^{2} = z^{2} - y^{2} q^{2}

Teile beide Seiten durch

p^{2}; p \neq = 0

x^{2} = \frac{z ^{2} - y ^{2} q ^{2}}{p ^{2}}; p \neq = 0

Für

x^{2} = f (a)

sind die Lösungen

x = f (a), - f (a)

x = \frac{z ^{2} - y ^{2} q ^{2}}{p ^{2}}, x = - \frac{z ^{2} - y ^{2} q ^{2}}{p ^{2}}; p \neq = 0

Vereinfache

\frac{z ^{2} - y ^{2} q ^{2}}{p ^{2}} : \frac{z ^{2} - y ^{2} q ^{2}}{p}

Vereinfache

- \frac{z ^{2} - y ^{2} q ^{2}}{p ^{2}} : - \frac{z ^{2} - y ^{2} q ^{2}}{p}

x = \frac{z ^{2} - y ^{2} q ^{2}}{p}, x = - \frac{z ^{2} - y ^{2} q ^{2}}{p}; p \neq = 0

(x + 2)^{3} = x^{3} - 8

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

\frac{( x ^{2} )}{6} - \frac{( 3 x )}{6} = 3 (x - 5)