erweitern (4y^2-1/(8y^3))^2

Lösung

erweitern

(4 y^{2} - \frac{1}{8 y ^{3}})^{2}

\frac{1024 y ^{10} - 64 y ^{5} + 1}{64 y ^{6}}

Schritte zur Lösung

(4 y^{2} - \frac{1}{8 y ^{3}})^{2}

Wende Formel für perfekte quadratische Gleichungen an:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(4 y^{2} - \frac{1}{8 y ^{3}})^{2} = (4 y^{2})^{2} - 2 \cdot 4 y^{2} \frac{1}{8 y ^{3}} + (\frac{1}{8 y ^{3}})^{2}

= (4 y^{2})^{2} - 2 \cdot 4 y^{2} \frac{1}{8 y ^{3}} + (\frac{1}{8 y ^{3}})^{2}

Vereinfache

(4 y^{2})^{2} - 2 \cdot 4 y^{2} \frac{1}{8 y ^{3}} + (\frac{1}{8 y ^{3}})^{2} : \frac{1024 y ^{10} - 64 y ^{5} + 1}{64 y ^{6}}

= \frac{1024 y ^{10} - 64 y ^{5} + 1}{64 y ^{6}}

\frac{d}{d p} ((t) (p + n^{2} a))

e x p an d (f \circ g) 3 (3 x^{2} + 7) + 5 (3 x^{2} + 7)

e x p an d x^{2} - \frac{5}{3} x

e x p an d \frac{2 ( x - 1 ) ^{2} ( x - 3 )}{( x + 2 ) ( x + 7 )}

e x p an d f^{'} (x) (45 - p (x))