✔ 最佳答案
Q: Simplify the following expression.
{[1-(1/n)]/n － (1+n)[1-(1/n)]} x 1/[n-(1/n)]
Sol:
{[1-(1/n)]/n － (1+n)[1-(1/n)]} x 1/[n-(1/n)]
＝ {[(n-1)/n]/n － (1+n)[(n-1)/n]} x 1/[(n²-1)/n]
＝ [(n-1)/n² － (n²-1)/n] x n/(n²-1)
＝ [(n-1)/n² － (n^3-n)/n²] x n/(n²-1)
＝ (n-1-n^3+n)/n² x n/(n²-1)
＝ - (n^3-2n-1)/n² x n/(n²-1)
＝ - (n^3-2n-1)/[n(n²-1)]
＝ - [(n+1)(n²-n-1)]/[n(n+1)(n-1)]
＝ - (n²-n-1)/(n²-n) ( 你可以只做到這個步驟 )
＝ - 1 ＋ 1/(n²-n)
Ans: - 1 ＋ 1/(n²-n)
2008-01-14 05:54:40 補充：
correction, from this step:＝ (n-1-n^3+n)/n² x n/(n²-1)＝ - (n^3-2n+1)/n² x n/(n²-1)＝ - (n^3-2n+1)/[n(n²-1)]＝ - [(n-1)(n²+n-1)]/[n(n+1)(n-1)]＝ - (n²+n-1)/(n²+n)Ans: - (n²+n-1)/(n²+n)
2008-01-14 05:56:00 補充：
correction, from this step:＝ (n-1-n^3+n)/n² x n/(n²-1)＝ - (n^3-2n+1)/n² x n/(n²-1)＝ - (n^3-2n+1)/[n(n²-1)]＝ - [(n-1)(n²+n-1)]/[n(n+1)(n-1)]＝ - (n²+n-1)/(n²+n) ( 你可以只做到這個步驟 )＝ - 1 ＋ 1/(n²+n)Ans: - 1 ＋ 1/(n²+n)
2008-01-14 06:01:45 補充：
I am sorry, I still get the different answer from yours.I try to substitute if n ＝ 2 to the questionit should be equal to - 5/6and substitute n ＝ 2 to my answerthe result is - 5/6
2008-01-14 06:02:11 補充：
but, substitute n ＝ 2 to your answerthe result is 2/3so, do you mind if check the question for me, please?
