<prompt> Acha x iwe nambari changamano kiasi kwamba x20111 na xneq 1. Kuhesabu jumlafracx2x-1 fracx4x2-1 fracx6x3-1 vidoti frac x4020x2010-1.  <response>: Acha S ionyeshe jumla iliyotolewa, kwa hivyoS fracx2x - 1 fracx4x2 - 1 vidoti fracx4020x2010 - 1 jumla_k 12010 fracx2kxk - 1. tag1Tunaweza kubadilisha mpangilio wa masharti, kupataS fracx4020x2010 - 1 fracx4018x2009 - 1 dots fracx 2x - 1 sum_k 12010 fracx4022 - 2kx2011 - k - 1.Tangu x2011 1,fracx4022 - 2kx2011 - k - 1 fracx-2kx-k - 1 frac1 xk - x2k frac1xk 1 - xk,hivyoS jumla_k 12010 frac1xk 1 - xk. tag2Kuongeza milinganyo 1 na 2, tunapataanzaalign2S sum_k 12010 fracx2kxk - 1 sum_k 12010 frac1x k 1 - xk sum_k 12010 kushoto fracx2kxk - 1 frac1xk 1 - xk haki sum_k 12010 kushoto fracx3kxk xk - 1 - frac1xk x k - 1 kulia jumla_k 12010 fracx3k - 1xk xk - 1.endalignTunaweza kubainisha x3k - 1 kama xk - 1x2k xk 1, hivyoanzaalign2S sum_k 12010 fracxk - 1x2k xk 1xk xk - 1 sum_k 12010 fracx2k xk 1xk sum_k 12010 kushoto xk 1 frac1xk kulia kushoto x 1 frac1x kulia kushoto x2 1 frac1x2 kulia vidoti kushoto x2010 1 frac1x2010 kulia x x2 dots x2010 2010 frac1x frac1x2 dots frac1101 x2010.endalignKwa kuwa x2011 1, tunayo x2011 - 1 0, ambayo huchangia kamax - 1x2010 x2009 dots x 1 0.Tunajua kwamba x neq 1, kwa hivyo tunaweza kugawanya pande zote mbili kwa x - 1, kupatax2010 x2009 vitone x 1 0.Kishaanzaalign2S x x2 dots x2010 2010 frac1x frac1x2 dots frac1 x2010 x x2 dots x2010 2010 fracx2010 x2009 dots xx2011 -1 2010 frac-11 2008,mwishoalignhivyo S 1004. 