erweitern (t^2+4t+4)*(2^{1/2}(t+2))

Lösung

erweitern

(t^{2} + 4 t + 4) \cdot (2^{\frac{1}{2}} (t + 2))

2^{\frac{1}{2}} t^{3} + 6 \cdot 2^{\frac{1}{2}} t^{2} + 12 \cdot 2^{\frac{1}{2}} t + 2^{\frac{1}{2}} \cdot 8

Schritte zur Lösung

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

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

Wende das Distributivgesetz an:

m (a + b + c) = ma + mb + m c

(2^{\frac{1}{2}} (t + 2)) (t^{2} + 4 t + 4) = 2^{\frac{1}{2}} (t + 2) t^{2} + 2^{\frac{1}{2}} (t + 2) \cdot 4 t + 2^{\frac{1}{2}} (t + 2) \cdot 4

= 2^{\frac{1}{2}} (t + 2) t^{2} + 2^{\frac{1}{2}} (t + 2) \cdot 4 t + 2^{\frac{1}{2}} (t + 2) \cdot 4

Schreibe

2^{\frac{1}{2}} (t + 2) t^{2} + 2^{\frac{1}{2}} (t + 2) \cdot 4 t + 2^{\frac{1}{2}} (t + 2) \cdot 4

um:

2^{\frac{1}{2}} t^{3} + 6 \cdot 2^{\frac{1}{2}} t^{2} + 12 \cdot 2^{\frac{1}{2}} t + 2^{\frac{1}{2}} \cdot 8

= 2^{\frac{1}{2}} t^{3} + 6 \cdot 2^{\frac{1}{2}} t^{2} + 12 \cdot 2^{\frac{1}{2}} t + 2^{\frac{1}{2}} \cdot 8

e x p an d \frac{( k + 1 ) ( k + 2 ) ( k + 3 )}{6}

s im pl i f y (p + \frac{a}{v ^{2}}) (v - b)

e x p an d cos ((5 - n) π)

e x p an d - \frac{4 x y + 8 y - 6}{( x + 2 ) ^{2}}

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