展开 (K(is+5))/((is(is+1)(is+2)))

解答

展开

\frac{K ( i s + 5 )}{( i s ( i s + 1 ) ( i s + 2 ) )}

\frac{- 13 K - K s ^{2}}{4 + s ^{4} + 5 s ^{2}} + i \frac{- 10 K + 2 K s ^{2}}{s ( 1 + s ^{2} ) ( 4 + s ^{2} )}

求解步骤

\frac{K ( i s + 5 )}{i s ( i s + 1 ) ( i s + 2 )}

乘开

i s (i s + 1) (i s + 2) : - i s^{3} - 3 s^{2} + 2 i s

= \frac{K ( i s + 5 )}{- i s ^{3} - 3 s ^{2} + 2 i s}

乘开

K (i s + 5) : i Ks + 5 K

= \frac{i Ks + 5 K}{- i s ^{3} - 3 s ^{2} + 2 i s}

\frac{i Ks + 5 K}{- i s ^{3} - 3 s ^{2} + 2 i s}

有理化:

\frac{- K s ^{4} - 13 K s ^{2} + i ( 2 K s ^{3} - 10 Ks )}{s ^{6} + 5 s ^{4} + 4 s ^{2}}

= \frac{- K s ^{4} - 13 K s ^{2} + i ( 2 K s ^{3} - 10 Ks )}{s ^{6} + 5 s ^{4} + 4 s ^{2}}

将

\frac{- K s ^{4} - 13 K s ^{2} + i ( 2 K s ^{3} - 10 Ks )}{s ^{6} + 5 s ^{4} + 4 s ^{2}}

改写成标准复数形式:

\frac{- K s ^{2} - 13 K}{s ^{4} + 5 s ^{2} + 4} + \frac{2 K s ^{2} - 10 K}{s ( s ^{2} + 1 ) ( s ^{2} + 4 )} i

= \frac{- K s ^{2} - 13 K}{s ^{4} + 5 s ^{2} + 4} + \frac{2 K s ^{2} - 10 K}{s ( s ^{2} + 1 ) ( s ^{2} + 4 )} i

e x p an d a (2, 3) b (- 5, - 8)

e x p an d (a a + 1) x^{3} - 3 x^{2} + 3 x - 1

e x p an d \frac{1}{y ( y + 1 )}

s im pl i f y (1 + z^{2}) (5 + \frac{4 - 4 z ^{2}}{1 + z ^{2}})

e x p an d (s - 3) (s - 2 + 3)