展开 (2pi+i2pin)(2pi-i2pin)

解答

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(2 π + i 2 πn) \cdot (2 π - i \cdot 2 πn)

4 π^{2} - 4 π^{2} i^{2} n^{2}

求解步骤

(2 π + i \cdot 2 πn) (2 π - i \cdot 2 πn)

= (2 π + 2 πin) (2 π - 2 πin)

使用平方差公式:

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

a = 2 π, b = i \cdot 2 πn

= (2 π)^{2} - (i \cdot 2 πn)^{2}

化简

(2 π)^{2} - (i \cdot 2 πn)^{2} : 4 π^{2} - 4 π^{2} i^{2} n^{2}

= 4 π^{2} - 4 π^{2} i^{2} n^{2}

e x p an d \frac{L ^{2} ( P + W L )}{16}

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

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

e x p an d \frac{( 2 x ^{2} - 6 x + 9 ) ^{2}}{2}

e x p an d (35) (67)