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

Lösung

erweitern

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

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

Schritte zur Lösung

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

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

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

Multipliziere aus

(x^{2} + 2 x + 1) (x + 3)^{'} : x^{2} (x + 3)^{'} + 2 x (x + 3)^{'} + (x + 3)^{'}

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

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

e x p an d \frac{2 ( 6 - 2 x ) ^{3}}{3}

e x p an d 2 (x + 9) - 18 ln (x + 9)

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

e x p an d \frac{x ^{4}}{( ( x - 1 ) ( x - 3 ) ^{3} )}