化简 (2+i/n)^2

解答

化简

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

\frac{4 n ^{2} - 1}{n ^{2}} + \frac{4}{n} i

求解步骤

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

使用完全平方公式:

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

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

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

化简

2^{2} + 2 \cdot 2 \cdot \frac{i}{n} + (\frac{i}{n})^{2} : \frac{4 n ^{2} - 1}{n ^{2}} + \frac{4}{n} i

= \frac{4 n ^{2} - 1}{n ^{2}} + \frac{4}{n} i

4 (v - 1) - 1 = 2 (2 v - 3)

s im pl i f y \frac{30}{8}

\frac{3}{4} (8 x - 4) = - 3

x^{2} + 6 x + 5 \geq 0

s im pl i f y (5 b^{3} + 9 b + 4) - (9 b - 4)