erweitern (e^{2x}-5)sech(x^2+7x-1)

Lösung

erweitern

(e^{2 x} - 5) sech (x^{2} + 7 x - 1)

e^{2 x} sech (x^{2} + 7 x - 1) - 5 sech (x^{2} + 7 x - 1)

Schritte zur Lösung

(e^{2 x} - 5) sech (x^{2} + 7 x - 1)

= sech (x^{2} + 7 x - 1) (e^{2 x} - 5)

Wende das Distributivgesetz an:

a (b - c) = ab - a c

a = sech (x^{2} + 7 x - 1), b = e^{2 x}, c = 5

= sech (x^{2} + 7 x - 1) e^{2 x} - sech (x^{2} + 7 x - 1) \cdot 5

= e^{2 x} sech (x^{2} + 7 x - 1) - 5 sech (x^{2} + 7 x - 1)

e x p an d l {(3 e^{- 3 t} - 2 t)^{2} + (cos (3 t) - e^{5 t})^{2}}

e x p an d \frac{( 2 x - 3 ) ( 1 - x ^{2} )}{( x ^{2} + 1 ) ( 5 - x )}

s im pl i f y (- 3 + i 3)^{14}

e x p an d (\frac{4}{16} x + \frac{5}{16} y + \frac{3}{16} z + \frac{4}{16} q)^{4}

e x p an d \frac{p + 2}{( p + 1 ) ( p - 2 ) ( p ^{2} + 4 )}