x^2+2x+1=4,

一个“整式的乘法”的问题请先阅读下列解题过程,再仿做下面的问题.已知X*X + X - 1=0,求X*X*X + 2*X*X + 3的值.解X*X*X + 2X*X +3=X*X*X +X*X -X +X*X +X +3=X{X*X +X -1} +X*X +X -1 +4=0+0+4=4+ x + X*X + X*X*X=0.+ X*X + X*X*X *-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X || (x/x-2)+(2/x*2+2x)-(x+6/x*2-4)化简,=x^2/x(x-2)+(2x-4)/x(x^2-4)-(x^2+6x)/x(x^2-4)=x^2/x(x-2)-(x+2)/x(x-2)=(x-2)(x+1)/x(x-2)=(x+1)/x (x-3/x-2)-(x-2/x-1)=(x-5/x4) -(x-4/x-3) x(x+1)+(x+2)+(x+3)+(x+4)+(x+5) = 27求x 解方程(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=1 X+X+X+(X-1)+(X-2)+(X-3)+(X-4)=10(X-3)+(X-4) x+2/x+1-x+4/x+3=x+6/x+5-x+8/x+7 x=? x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002 写过程x(x-1)+2x(x+1)-3x(2x-5)-x(3x-2)+2x(2-x)=-5x(x-2)-4 1.用公式计算：COSX=1-X*X/2!X*X*X*X/4!-X*X*X*X*X*X/6!.直到最后一项的绝对 1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x*x)+8/(1-x*x*x*x*x*x*x*x)怎么做 (x^2+x-1)(x^2+x+2)=4 （X+4)(x-2)(x-4),其中x=-1 (x-2)(x+1)(x+4)(x+7)=19 (x+1)(x+2)(x+3)(x+4)=24 (x-1)(x-2)(x-3)(x-4)=48 (x+1)(x+2)(x+3)(x+4)=24