1-b分之1+1+b分之1+1+b²+1+b四次方（逐步通分）

1-b分之1+1+b分之1+1+b²分之1+1+b四次方分之1
=1/(1-b)+1/(1+b)+2/（1+b²)+4/(1+b⁴)
=(1+b+1-b)/(1-b²)+2/(1+b²)+4/(1+b²)
=(2+2b²+2-2b²)/(1-b⁴)+4/(1+b⁴)
=(4+4b⁴+4-4b⁴)/(1+b^8)
=8/(1+b^8)

(y²+3y+2)分之1+(y²+5y+6)分之1+(y²+7y+12)分之1
=1/(y+1)(y+2)+1/(y+2)(y+3)+1/(y+3)(y+4)
=1/(y+1)-1/(y+2)+1/(y+2)-1/(y+3)+1/(y+3)-1/(y+4)
=1/(y+1)-1/(y+4)
=(y+4-y-1)/(y+1)(y+4)
=3/(y²+5y+4)