化简 (8+i/n)^2

解答

化简

(8 + \frac{i}{n})^{2}

\frac{64 n ^{2} - 1}{n ^{2}} + \frac{16}{n} i

求解步骤

(8 + \frac{i}{n})^{2}

使用完全平方公式:

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

(8 + \frac{i}{n})^{2} = 8^{2} + 2 \cdot 8 \cdot \frac{i}{n} + (\frac{i}{n})^{2}

= 8^{2} + 2 \cdot 8 \cdot \frac{i}{n} + (\frac{i}{n})^{2}

化简

8^{2} + 2 \cdot 8 \cdot \frac{i}{n} + (\frac{i}{n})^{2} : \frac{64 n ^{2} - 1}{n ^{2}} + \frac{16}{n} i

= \frac{64 n ^{2} - 1}{n ^{2}} + \frac{16}{n} i

e x p an d e^{2 x} (4 e^{3 x} - 3 e^{- 3 x})

e x p an d 4 - x^{2} \cdot 1 + x

s im pl i f y (m + m - 3)^{2}

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

e x p an d 4 x^{2} - x (8 + 9 x^{4} - 5 x)