erweitern I(R+(2piFL-1/(2piFC)))

Lösung

erweitern

I (R + (2 π F L - \frac{1}{2 π FC}))

R I + 2 π L I F - \frac{I}{2 π FC}

Schritte zur Lösung

I (R + (2 π F L - \frac{1}{2 π FC}))

= I (R + (2 π L F - \frac{1}{2 π FC}))

Wende das Distributivgesetz an:

a (b + c) = ab + a c

a = I, b = R, c = (2 π F L - \frac{1}{2 π FC})

= I R + I (2 π F L - \frac{1}{2 π FC})

= R I + I (2 π L F - \frac{1}{2 π FC})

Multipliziere aus

I (2 π L F - \frac{1}{2 π FC}) : 2 π L I F - \frac{I}{2 π FC}

= R I + 2 π L I F - \frac{I}{2 π FC}

e x p an d \frac{- m + 3}{3 ( - m + 9 )}

s im pl i f y (4 x + 2 \circ 6 x) (- 3)

e x p an d \frac{1}{( 1 + 2 x ) ( 1 + x ^{2} )}

e x p an d (2 x + 3) e^{x}

e x p an d (\frac{3 , 5 x ^{2}}{12} + 20 x) (\frac{3.5 x}{6})