作业帮1/1x2+1/2x3+1/3x4+...+1/2007x2008=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/08 22:44:20
1/1x2+1/2x3+1/3x4+...+1/2007x2008=?
1/1x2+1/2x3+1/3x4+...+1/2007x2008=?
1/1x2+1/2x3+1/3x4+...+1/2007x2008=?
1/1x2+1/2x3+1/3x4+...+1/2007x2008
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2007-1/2008)
=1-1/2+1/2-1/3+1/3-1/4+……+1/2007-1/2008
中间互相抵消
=1-1/2008
=2007/2008
1/1x2=1/1-1/2
1/2x3=1/2-1/3
.....
1/1x2+1/2x3+1/3x4+...+1/2007x2008
=1-1/2+1/2-1/3+.....+1/2007-1/2008
=1-1/2008
=2007/2008
原式=(1-1/2)+(1/2-1/3)…………+(1/2007+1/2008)
=1-1/2+1/2-1/3+1/3…………+1/2007-1/2008
=1-1/2008
=2007/2008
因为1/1×2=1-1/2
1/2×3=1/2-1/3
……
1/2007×2008=1/2007-1/2008
所以原式=1-1/2+1/2-1/3+1/3-1/4+……+1/2007-1/2008
=1-1/2008=2007/2008
是不是1*2、2*3……都是分母呀，否则真的不好做
用克拉默法则解下列方程组 x1-2x2+3x3-4x4=4 x2-x3+x4=-3 x1+3x2+2x4=1 -7x2+3x3+x4=3x1-2x2+3x3-4x4=4x2-x3+x4=-3 x1+3x2+2x4=1 -7x2+3x3+x4=3
因式分解：x4+x3+2x2 +x+1
解方程组X1-2x2+3x3-x4=1,3x1-x2+5x3-3x4=2,2x1+x2+2x3-2x4=3
x1+5x2-x3-x4=-1x1-2x2+x3+3x4=33x1+8x2-x3+x4=1
写出方程组2*x1+x2-x3+x4=1,x1+2*x2+x3-x4=2,x1+x2+2*x3+x4=3的通解?
求非齐次线方程组的通解 ：2x1+x2-x3+x4=1 x1+2x2+x3-x4=2 x1+x2+2x3+x4=3
具体写出方程组：2x1+x2-x3+x4=1;x1+2x2+x3-x4=2;x1+x2+2x3+x4=3的通解
解一道方程组x1+x2+x3=5,x2+x3+x4=1,x3+x4+x5=-5,x4+x5+x1=-3,x5+x1+x2=2
用初等行变换来解下列线性方程组（1）2x1-x2+3x3=3 3x1+x2-5x3=0 4x1-x2+x3=3 x1+3x2-13x3=-6(2) x1-2x2+x3+x4=1 x1-2x2+x3-x4=-1 x1-2x2+x3-5x4=5(3) x1-x2+x3-x4=1 x1-x2-x3+x4=0 x1-x2-2x3+2x4=-1/2
1X2+2X3+3X4+.NX(N+1)
1X2+2X3+3X4+.NX(N+1)规律
1x2+2x3+3x4+.+2011x2012
1x2+2x3+3x4+4x5+.+15x16
1x2+2x3+3x4+.+100x101=
1x2+2x3+3x4+.+100x101=?结果
1x1!+2x2!+3x3!+4x4!.nxn!
1X2+2x3+3x4+……+2013x2014
1x2+2x3+3x4+.+19x20