Question

(x²+x)分之一+(x²+3x+2)分之1+(x²+5x+6)分之1+(x²+7x+12)分之1

Answer 1

(x²+x)分之一 = 1/x - 1/(x+1)
(x²+3x+2)分之1=1/(x+1) - 1/(x+2)
(x²+5x+6)分之1=1/(x+2) - 1/(x+3)
(x²+7x+12)分之1=1/(x+3) - 1/(x+4)

右边相加 = 1/x- 1/(x+4)

Answer 2

1/(x²+x)+1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)

=1/[x(x+1)]+1/[(x+1)(x+2)]+1/(x+2)(x+3)]+1/[(x+3)(x+4)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
=(x+4-x)/[x(x+4)]
=4/(x²+4x)

Answer 3

原式=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)

=1/x-1/(x+4)
=4/(x²+4x)

Answer 4

=1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]1/
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
=4/[x(x+4)]