erweitern ((2n^2+n+1)^2)/4

Lösung

erweitern

\frac{( 2 n ^{2} + n + 1 ) ^{2}}{4}

n^{4} + n^{3} + \frac{5 n ^{2}}{4} + \frac{n}{2} + \frac{1}{4}

Schritte zur Lösung

\frac{( 2 n ^{2} + n + 1 ) ^{2}}{4}

(2 n^{2} + n + 1)^{2} = 4 n^{4} + 4 n^{3} + 5 n^{2} + 2 n + 1

= \frac{4 n ^{4} + 4 n ^{3} + 5 n ^{2} + 2 n + 1}{4}

Wende Bruchregel an:

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

\frac{4 n ^{4} + 4 n ^{3} + 5 n ^{2} + 2 n + 1}{4} = \frac{4 n ^{4}}{4} + \frac{4 n ^{3}}{4} + \frac{5 n ^{2}}{4} + \frac{2 n}{4} + \frac{1}{4}

= \frac{4 n ^{4}}{4} + \frac{4 n ^{3}}{4} + \frac{5 n ^{2}}{4} + \frac{2 n}{4} + \frac{1}{4}

Teile die Zahlen:

\frac{4}{4} = 1

= n^{4} + n^{3} + \frac{5 n ^{2}}{4} + \frac{2 n}{4} + \frac{1}{4}

Streiche

\frac{2 n}{4} : \frac{n}{2}

= n^{4} + n^{3} + \frac{5 n ^{2}}{4} + \frac{n}{2} + \frac{1}{4}

e x p an d \frac{1}{( s + a ) \cdot ( s + b ) ^{2}}

e x p an d \frac{1}{5 ( x + 1 )}

e x p an d (2 n + 1) π - \frac{π}{2}

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

e x p an d (2 T - 1) (π + 5)