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

Lösung

erweitern

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

\frac{x ^{2} + 6 x + 9}{x ^{5} + 7 x ^{4} + 19 x ^{3} + 25 x ^{2} + 16 x + 4}

Schritte zur Lösung

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

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

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

(x + 2)^{2} = x^{2} + 4 x + 4

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

(x + 3)^{2} = x^{2} + 6 x + 9

= \frac{x ^{2} + 6 x + 9}{( x ^{3} + 3 x ^{2} + 3 x + 1 ) ( x ^{2} + 4 x + 4 )}

Multipliziere aus

(x^{3} + 3 x^{2} + 3 x + 1) (x^{2} + 4 x + 4) : x^{5} + 7 x^{4} + 19 x^{3} + 25 x^{2} + 16 x + 4

= \frac{x ^{2} + 6 x + 9}{x ^{5} + 7 x ^{4} + 19 x ^{3} + 25 x ^{2} + 16 x + 4}

e x p an d (f (x) = 2, x + 5) \cdot (g (x) = x + 5)

e x p an d - 3 (3 x - 5 = 7 - 4 x)

e x p an d (4 + 6 x) x^{3} e^{6 x}

s im pl i f y 4 \cdot (1 - \frac{cos ( 2 θ )}{2})

s im pl i f y (x - 14)^{2}