1⼀(x^2+x)+1⼀(x^2+3x+2)+1⼀(x^2+5x+6)+1⼀(x^2+7x+12)

额。。。好长的题目。
2025-01-19 08:09:44
推荐回答(2个)
回答1:

1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+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)
=4/(x^2+4x).

回答2:

=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/(x^2+4x).