[b/(a^2-ab)+a/(b^2-ab)]*ab/(a+b)=?
已知1/(x-2)-1/(x-1)=1/(x-1)(x-2),
1/(x-3)-1/(x-2)=1/(x-2)(x-3),
则1/(x-1)+1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)=?
[b/(a^2-ab)+a/(b^2-ab)]*ab/(a+b)=?
已知1/(x-2)-1/(x-1)=1/(x-1)(x-2),
1/(x-3)-1/(x-2)=1/(x-2)(x-3),
则1/(x-1)+1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)=?
第一题:原式=[b^2/ab(a-b)-a^2/ab(a-b)]*ab/(a+b)=(b^2-a^2)/(a^2-b^2)=-1
第二题:原式=1/(x-1)+1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+...+1/(x-100)-1/(x-99)=1/(x-100)
第二题这个求和的和式里面前面198项两两抵消了。所以最后就等于最后一项展开以后未被抵消的一部分。
[b/(a^2-ab)+a/(b^2-ab)]*ab/(a+b)=b^2/(a^2-b^2)+a^2/(b^2-a^2)=-1
1/(x-1)+1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)=1/(x-100)
能用语言说说么?
你打的这个实在是……
不好看