2013!-(1!+2!x2+3!+4!x4+.+2011!x2011+2012!x2012)=
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:2024/04/28 05:42:46
2013!-(1!+2!x2+3!+4!x4+.+2011!x2011+2012!x2012)=
2013!-(1!+2!x2+3!x3+4!x4+.+2011!x2011+2012!x2012) (1)
=2013!-1!-2!x2-3!x3-4!x4-.-2011!x2011-2012!x2012
=2013!-2012!x(2013-1)-2011!x2011-.-4!x4-3!x3-2!x2-1!x1
=2013!-2013!+2012!-2011!x2011-.-4!x4-3!x3-2!x2-1!x1
=2012!-2011!x2011-.-4!x4-3!x3-2!x2-1!x1 (2)
=2012!-2011!x(2012-1)-2010!x2010-.-4!x4-3!x3-2!x2-1!x1
=2011!-2010!x2010-.-4!x4-3!x3-2!x2-1!x1 (3)
=2010!-2009!x2009-.-4!x4-3!x3-2!x2-1!x1 (4 ) (从1、2、3推出4)
.
=3!-2!x2-1!x1
=2!-1!x1
=1
过程、结果如上.
=2013!-1!-2!x2-3!x3-4!x4-.-2011!x2011-2012!x2012
=2013!-2012!x(2013-1)-2011!x2011-.-4!x4-3!x3-2!x2-1!x1
=2013!-2013!+2012!-2011!x2011-.-4!x4-3!x3-2!x2-1!x1
=2012!-2011!x2011-.-4!x4-3!x3-2!x2-1!x1 (2)
=2012!-2011!x(2012-1)-2010!x2010-.-4!x4-3!x3-2!x2-1!x1
=2011!-2010!x2010-.-4!x4-3!x3-2!x2-1!x1 (3)
=2010!-2009!x2009-.-4!x4-3!x3-2!x2-1!x1 (4 ) (从1、2、3推出4)
.
=3!-2!x2-1!x1
=2!-1!x1
=1
过程、结果如上.
1/1x2+1/2x3+1/3x4+.+1/2011x2012=?
已知:x1-1的绝对值+(x2-2)的平方+x3-3的绝对值的立方+(x4-4)的4次方+.+(x2011-2011)2
1x2分之一+2x3分之一+3x4分之一+.+2010x2011分之一等于多少
2x4分之1+4x6分之1……2010x2012分之1=?为什么
已知/X1-1/+/X2-2/+/X3-3/+...+/X2010-2010/+/X2011-2011/=0,试求代数式
2X4分之一+4X6分之一+6X8分之一+…+2010X2012分之一= n(n+1)分之一=
1/2x4+1/4x6+1/6x8+…+1/2010x2012 怎么算!
1/2x4+1/4x6+1/6x8+.+1/2010x2012
求齐次线性方程组x1+2x2+x3+x4+x5=1 2x1+4x2+3x3+x4+x5=2 -x1-2x2+x3+3x4
1分解因式:2x4-x3-13x2-15=?2.分解因式:x5+x4+x3+x2+x+1=?3分解因式:x4-4x2+6
解下列方程组 x1+x2=x2+x3=...=x2012+x2013=1 x1+x2+...+x2013=2013
有理数的计算(1)1/1x4+1/4x7+1/7x10+...+1/2008x2011+1/2011x2014