搜搜做题作业网计算1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+...1/(x+2004)(x+2006)
来源：学生作业帮 编辑：搜搜做题作业网作业帮 分类：数学作业 时间：2024/07/13 19:15:04
计算1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+...1/(x+2004)(x+2006)
1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+...1/(x+2004)(x+2006)
=1/x(x+2)+{1/(x+2)-1/(x+4)+1/(x+4)-1/(1+6)+...+1/(x+2004)-1/(x+2006)}/2
=1/x-1/(x+2)+{1/(x+2)-1/(x+2006)}/2
=1/x-1/(2x+4)-1/(2x+4012)
=1/x(x+2)+{1/(x+2)-1/(x+4)+1/(x+4)-1/(1+6)+...+1/(x+2004)-1/(x+2006)}/2
=1/x-1/(x+2)+{1/(x+2)-1/(x+2006)}/2
=1/x-1/(2x+4)-1/(2x+4012)
计算（x*x+4x+5）/（x+2）-（x*x+6x+10）/（x+3）+1
x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002
|X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10|
如何计算 (x+1)(x+2)(x+3)(x+4)
计算(X-1)(X-2)(X-3)(X-4)
x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4
x^5+x^4 = (x^3-x)(x^2+x+1)+x^2+x
(x-1)(x^2+x+1)-x(x^2-2x+4)计算
分式计算：(x+2)/(x^2-2x)-(x-1)/(x^2-4x+4)
计算:2x/(x-2)(x+1)÷(1－x/x+1)-x²+2x/x²-4?
x+2/x²-2x-x-1/x²-4x+4计算
计算：(x-1)(x^2-4X+3)-x(x^2-5X+2)有过程