faktorisieren 5t(6t+4)^{(-1)/2}+6(6t+4)^{1/2}

Lösung

faktorisieren

5 t (6 t + 4)^{\frac{- 1}{2}} + 6 (6 t + 4)^{\frac{1}{2}}

\frac{41 t + 24}{2 ^{\frac{1}{2}} ( 3 t + 2 ) ^{\frac{1}{2}}}

Schritte zur Lösung

5 t (6 t + 4)^{\frac{- 1}{2}} + 6 (6 t + 4)^{\frac{1}{2}}

Wende Exponentenregel an:

a^{b + c} = a^{b} a^{c}

(6 t + 4)^{\frac{1}{2}} = (6 t + 4)^{1 + \frac{- 1}{2}} = (6 t + 4)^{1} (6 t + 4)^{\frac{- 1}{2}}

= 5 t (6 t + 4)^{\frac{- 1}{2}} + 6 (6 t + 4)^{1} (6 t + 4)^{\frac{- 1}{2}}

Klammere gleiche Terme aus

(4 + 6 t)^{\frac{- 1}{2}}

= (4 + 6 t)^{\frac{- 1}{2}} (5 t + 6 (4 + 6 t))

Faktorisiere

5 t + 6 (6 t + 4) : 41 t + 24

= (41 t + 24) (6 t + 4)^{\frac{- 1}{2}}

Fasse zusammen

= \frac{41 t + 24}{( 6 t + 4 ) ^{\frac{1}{2}}}

(4 + 6 t)^{\frac{1}{2}} = (2 (3 t + 2))^{\frac{1}{2}}

= \frac{41 t + 24}{( 2 ( 3 t + 2 ) ) ^{\frac{1}{2}}}

(2 (3 t + 2))^{\frac{1}{2}} = 2^{\frac{1}{2}} (3 t + 2)^{\frac{1}{2}}

= \frac{41 t + 24}{2 ^{\frac{1}{2}} ( 3 t + 2 ) ^{\frac{1}{2}}}

s im pl i f y \frac{- 6}{- 3}

0.22 m = 49.95 + 0.08 m

6 x^{2} - 3 x + 4 = 8 x + 1

f a c t or 4 x^{2} - 13 x - 12

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