erweitern (6x+1)/((x+1)^3(x^2+7)^2)

Lösung

erweitern

\frac{6 x + 1}{( x + 1 ) ^{3} ( x ^{2} + 7 ) ^{2}}

\frac{6 x + 1}{x ^{7} + 3 x ^{6} + 17 x ^{5} + 43 x ^{4} + 91 x ^{3} + 161 x ^{2} + 147 x + 49}

Schritte zur Lösung

\frac{6 x + 1}{( x + 1 ) ^{3} ( x ^{2} + 7 ) ^{2}}

(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1

= \frac{6 x + 1}{( x ^{3} + 3 x ^{2} + 3 x + 1 ) ( x ^{2} + 7 ) ^{2}}

(x^{2} + 7)^{2} = x^{4} + 14 x^{2} + 49

= \frac{6 x + 1}{( x ^{3} + 3 x ^{2} + 3 x + 1 ) ( x ^{4} + 14 x ^{2} + 49 )}

Multipliziere aus

(x^{3} + 3 x^{2} + 3 x + 1) (x^{4} + 14 x^{2} + 49) : x^{7} + 3 x^{6} + 17 x^{5} + 43 x^{4} + 91 x^{3} + 161 x^{2} + 147 x + 49

= \frac{6 x + 1}{x ^{7} + 3 x ^{6} + 17 x ^{5} + 43 x ^{4} + 91 x ^{3} + 161 x ^{2} + 147 x + 49}

e x p an d (15 - t^{2}) e^{t}

e x p an d \frac{5 ( x ^{2} ( 2 x ^{2} + 3 ) - 36 )}{12 x ^{4}}

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

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

s im pl i f y 2 \cdot δ (x - 4) + 7 H (x - 11)