erweitern ((2x+1)(x^2-x-4))/(2x)

Lösung

erweitern

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

x^{2} - \frac{x}{2} - \frac{9}{2} - \frac{2}{x}

Schritte zur Lösung

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

Multipliziere aus

(2 x + 1) (x^{2} - x - 4) : 2 x^{3} - x^{2} - 9 x - 4

= \frac{2 x ^{3} - x ^{2} - 9 x - 4}{2 x}

Wende Bruchregel an:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 x ^{3} - x ^{2} - 9 x - 4}{2 x} = \frac{2 x ^{3}}{2 x} - \frac{x ^{2}}{2 x} - \frac{9 x}{2 x} - \frac{4}{2 x}

= \frac{2 x ^{3}}{2 x} - \frac{x ^{2}}{2 x} - \frac{9 x}{2 x} - \frac{4}{2 x}

\frac{2 x ^{3}}{2 x} = x^{2}

\frac{x ^{2}}{2 x} = \frac{x}{2}

\frac{9 x}{2 x} = \frac{9}{2}

\frac{4}{2 x} = \frac{2}{x}

= x^{2} - \frac{x}{2} - \frac{9}{2} - \frac{2}{x}

e x p an d \frac{2 x ^{3} + 5 x ^{2} + 4 x + 3}{( x + 1 ) ^{2}}

e x p an d [\frac{7}{( x + 4 ) ^{2}}]

e x p an d ((x + 1) (x - 1))^{3}

e x p an d \frac{( x ^{2} - 2 x - 15 ) ( x + 3 )}{( - x ^{2} - 1 )}

s im pl i f y (3 x^{2} - 5 x - 4 \circ 2 x + 2) (3)