1樓：匿名使用者
解：1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1(x+2009)(x+2010)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2009)-1/(x+2010)
=1/x-1/(x+2010)
故當x=1時，原式=1 -1/2011 =2010/2011
2樓：三味學堂答疑室
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+……+1/(x+2009)-1/(x+2010)
=1/x-1/(x+2010）
當x=1時，原式=1-1/2011=2010/2011
3樓：韻淵
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+2009)(x+2010)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+2009)-1/(x+2010)
=1/x-1/(x+2010)
=2010/x(x+2010)
當x=1時，原式=2010/2011
X 2 1 X 1 9 a0 a1x a2x 2a11 x11,求展開式中係數最大項
1 設 x 1 9 b0 b1x b2x 2 b9x 9 x 2 x 1 9 b0x 2 b1x 3 b9x 11 上二式相加得 x 2 1 x 1 9 b0 b1x b0 b2 x 2 b1 b3 x 3 b7 b9 x 9 b8x 8 b9x 11 b0 c 0,9 b1 c 1,9 b2 c ...
x2x1x4x3x,x2x1x4x3x3x2x5x4,初二數學計算題
原式 1 1 x 1 1 1 x 3 1 1 x 2 1 1 x 4 1 x 1 1 x 3 1 x 2 1 x 4 1 x 1 1 x 4 1 x 2 1 x 3 2x 5 x 1 x 4 2x 5 x 2 x 3 2x 5 1 x 1 x 4 1 x 2 x 3 2x 5 x 2 5x 6 x ...
x 2x 2 2x 1x 2x 2 1x 1）,其中x（根號
1 baix 2 du x 2 2x 1 zhix 2 daox 2 1 x 1 其中x 根號 回2 2 1 x 2 x 1 2 x 2 x 1 x 1 x 1 1 x 2 x 1 x 2 1 x 2 x 1 x 2 x 2 x 2 x 1 x 2 答x 2 x 2 x 2 3x 2 x 2 x 2...