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受欢迎的代数 >

展开 (2xe^{2x}+e^{2x}*2x^2)^6

解答

展开

(2 x e^{2 x} + e^{2 x} \cdot 2 x^{2})^{6}

解答

64 e^{12 x} x^{6} + 384 e^{12 x} x^{7} + 960 e^{12 x} x^{8} + 1280 e^{12 x} x^{9} + 960 e^{12 x} x^{10} + 384 e^{12 x} x^{11} + 64 e^{12 x} x^{12}

求解步骤

(2 x e^{2 x} + e^{2 x} \cdot 2 x^{2})^{6}

= (2 e^{2 x} x + 2 e^{2 x} x^{2})^{6}

使用二项式定理:

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

a = 2 x e^{2 x}, b = e^{2 x} \cdot 2 x^{2}

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

展开求和

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

化简

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

化简

\frac{6 !}{1 ! ( 6 - 1 ) !} (2 x e^{2 x})^{5} (e^{2 x} \cdot 2 x^{2})^{1} : 384 e^{12 x} x^{7}

化简

\frac{6 !}{2 ! ( 6 - 2 ) !} (2 x e^{2 x})^{4} (e^{2 x} \cdot 2 x^{2})^{2} : 960 e^{12 x} x^{8}

化简

\frac{6 !}{3 ! ( 6 - 3 ) !} (2 x e^{2 x})^{3} (e^{2 x} \cdot 2 x^{2})^{3} : 1280 e^{12 x} x^{9}

化简

\frac{6 !}{4 ! ( 6 - 4 ) !} (2 x e^{2 x})^{2} (e^{2 x} \cdot 2 x^{2})^{4} : 960 e^{12 x} x^{10}

化简

\frac{6 !}{5 ! ( 6 - 5 ) !} (2 x e^{2 x})^{1} (e^{2 x} \cdot 2 x^{2})^{5} : 384 e^{12 x} x^{11}

化简

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

= 64 e^{12 x} x^{6} + 384 e^{12 x} x^{7} + 960 e^{12 x} x^{8} + 1280 e^{12 x} x^{9} + 960 e^{12 x} x^{10} + 384 e^{12 x} x^{11} + 64 e^{12 x} x^{12}

流行的例子

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

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

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e x p an d \frac{- 50}{( s + 0.5 ) ( 0.2 s + 1 )}

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

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e x p an d (e^{t} + 1) h (t - 2)