展开 (4/x-1/(x^3))^2

解答

展开

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

\frac{16 x ^{4} - 8 x ^{2} + 1}{x ^{6}}

求解步骤

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

使用完全平方公式:

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

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

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

化简

(\frac{4}{x})^{2} - 2 \cdot \frac{4}{x} \cdot \frac{1}{x ^{3}} + (\frac{1}{x ^{3}})^{2} : \frac{16 x ^{4} - 8 x ^{2} + 1}{x ^{6}}

= \frac{16 x ^{4} - 8 x ^{2} + 1}{x ^{6}}

e x p an d (\frac{1}{20} e^{\frac{x}{20}} - \frac{1}{20} e^{- \frac{x}{20}})^{2}

e x p an d A (- 2, - 3) B (x_{2}, 2)

s im pl i f y (I - E D) (I + \frac{E D}{I - D E})

e x p an d cos (π t) - cos (π t) u (t - 2)

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