展开-(lk(lsin((2pix)/l)+2Cpi))/(4pi^2)

解答

展开

- \frac{l k ( l sin ( \frac{2 π x}{l} ) + 2 C π )}{4 π ^{2}}

- \frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}} - \frac{lC k}{2 π}

求解步骤

- \frac{l k ( l sin ( \frac{2 π x}{l} ) + 2 C π )}{4 π ^{2}}

乘开

l k (l sin (\frac{2 π x}{l}) + 2 C π) : l^{2} k sin (\frac{2 π x}{l}) + 2 π lC k

= - \frac{l ^{2} k sin ( \frac{2 π x}{l} ) + 2 π lC k}{4 π ^{2}}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{l ^{2} k sin ( \frac{2 π x}{l} ) + 2 π lC k}{4 π ^{2}} = - (\frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}}) - (\frac{2 π lC k}{4 π ^{2}})

= - (\frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}}) - (\frac{2 π lC k}{4 π ^{2}})

去除括号:

(a) = a

= - \frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}} - \frac{2 π lC k}{4 π ^{2}}

消掉

\frac{2 π lC k}{4 π ^{2}} : \frac{lC k}{2 π}

= - \frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}} - \frac{lC k}{2 π}

e x p an d 3 (y - 40 = - 38)

e x p an d f (∣ x ∣ - 1) e^{∣ x ∣}

e x p an d 1 - x ((- 2 - x) (4 - x))

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

e x p an d 2 x 2 x^{1}