erweitern (2x^2+1)^6

Lösung

erweitern

(2 x^{2} + 1)^{6}

64 x^{12} + 192 x^{10} + 240 x^{8} + 160 x^{6} + 60 x^{4} + 12 x^{2} + 1

Schritte zur Lösung

(2 x^{2} + 1)^{6}

Wende Binomialsetz an:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 x^{2}, b = 1

= i = 0 \sum 6 (i 6) (2 x^{2})^{(6 - i)} \cdot 1^{i}

Erweitere Summation

= \frac{6 !}{0 ! ( 6 - 0 ) !} (2 x^{2})^{6} \cdot 1^{0} + \frac{6 !}{1 ! ( 6 - 1 ) !} (2 x^{2})^{5} \cdot 1^{1} + \frac{6 !}{2 ! ( 6 - 2 ) !} (2 x^{2})^{4} \cdot 1^{2} + \frac{6 !}{3 ! ( 6 - 3 ) !} (2 x^{2})^{3} \cdot 1^{3} + \frac{6 !}{4 ! ( 6 - 4 ) !} (2 x^{2})^{2} \cdot 1^{4} + \frac{6 !}{5 ! ( 6 - 5 ) !} (2 x^{2})^{1} \cdot 1^{5} + \frac{6 !}{6 ! ( 6 - 6 ) !} (2 x^{2})^{0} \cdot 1^{6}

Vereinfache

\frac{6 !}{0 ! ( 6 - 0 ) !} (2 x^{2})^{6} \cdot 1^{0} : 64 x^{12}

Vereinfache

\frac{6 !}{1 ! ( 6 - 1 ) !} (2 x^{2})^{5} \cdot 1^{1} : 192 x^{10}

Vereinfache

\frac{6 !}{2 ! ( 6 - 2 ) !} (2 x^{2})^{4} \cdot 1^{2} : 240 x^{8}

Vereinfache

\frac{6 !}{3 ! ( 6 - 3 ) !} (2 x^{2})^{3} \cdot 1^{3} : 160 x^{6}

Vereinfache

\frac{6 !}{4 ! ( 6 - 4 ) !} (2 x^{2})^{2} \cdot 1^{4} : 60 x^{4}

Vereinfache

\frac{6 !}{5 ! ( 6 - 5 ) !} (2 x^{2})^{1} \cdot 1^{5} : 12 x^{2}

Vereinfache

\frac{6 !}{6 ! ( 6 - 6 ) !} (2 x^{2})^{0} \cdot 1^{6} : 1

= 64 x^{12} + 192 x^{10} + 240 x^{8} + 160 x^{6} + 60 x^{4} + 12 x^{2} + 1

f (x) = (\frac{x - 1}{2})^{\frac{1}{4}} - 13 + 5 x

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

s im pl i f y \frac{( 2 x ^{3} - 14 x )}{( 5 x ^{2} + 35 x )}

s im pl i f y (\frac{x - 1}{3})^{3}

f a c t or 6 x^{2} y + 12 x^{3} y^{2} - 3 x^{2} y^{2}