2001x(8/1-1/2009)+8x(1/2001-1/2009)-2009x(1/8+1/2009)+8 怎么用简

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/05/18 23:38:25

2001x(8/1-1/2009)+8x(1/2001-1/2009)-2009x(1/8+1/2009)+8 怎么用简便方法计算

应该是2001x(1/8-1/2009)+8x(1/2001-1/2009)-2009x(1/8+1/2009)+8
=2001/8-2001/2009+8/2001-8/2009-2009/8-2009/2009+8
=（2001/8-2009/8）-（2001/2009+8/2009）+8/2001-1+8
=-1-1+8/2001-1+8=5又8/2001
再问：谢谢啦
再答：不客气

|X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10| 3x-1>x+1 { x+8 解方程(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=1 x^8+x^6+x^4+x^2+1因式分解 x/2+x/8+x/12+x/20=1 X-5+8*X=7*X-2*X-1 因式分解x^8+x^6+x^4+x^2+1 计算(x+/1x)(x^2+1/x^2)(x^4+1/x^4)(x^8+1/x^8)(x^16+1/x^16)(x^2- [1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)] 解方程：1/(x*x+11x-8)+1/(x*x+-8)+1/(x*x-3x-8)=0 化简：1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+8)(x+9) y=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)的导数在x=1