Question

y平方加3y加2分之1加y平方加5y加6分之1加y平方加7y加12用列项消除法做此题

Answer 1

题目因该是 1/(Y^2+3Y+2)+1/(Y^2+5Y+6)+1/(Y^2+7Y+12)

=1/(Y+1)(Y+2)+1/(Y+2)(Y+3)+1/(Y+3)(Y+4)

=[1/(Y+1)-1/(Y+2)]+1/[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/[(Y+1)(Y+4)]