faktorisieren 9b^{2x}-1/4

Lösung

faktorisieren

9 b^{2 x} - \frac{1}{4}

(3 b^{x} + \frac{1}{2}) (3 b^{x} - \frac{1}{2})

Schritte zur Lösung

9 b^{2 x} - \frac{1}{4}

Schreibe

9 b^{2 x}

um:

(3 b^{1 x})^{2}

= (3 b^{1 \cdot x})^{2} - \frac{1}{4}

Wende Radikal Regel an:

a = (a)^{2}

\frac{1}{4} = (\frac{1}{4})^{2}

= (3 b^{1 x})^{2} - (\frac{1}{4})^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

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

= (3 b^{1 x} + \frac{1}{4}) (3 b^{1 x} - \frac{1}{4})

Fasse zusammen

= (3 b^{x} + \frac{1}{2}) (3 b^{x} - \frac{1}{2})

s im pl i f y e^{2 + 2 i}

f a c t or [(x + 8) - y]^{2}

f a c t or x (a + b - c) - (- y (a + b - c))

f a c t or 21 b + 63 b

f (x) = [x^{2}]