(1.)3X^2-4=0.(2)(x-1)^2=2

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 已知x=2π+1.求(X-1)+(X-2)+(X-3)+(X-4)+(X-5)+(X-6)+(X-7)+(X-8)+(X-9)+(X-10) *-----------------------------------------------*| 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-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 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 已知1+x+x^2+x^3+x^4+.+x^4=0.求多项式1+x+x^2+x^3+.+x^1989 1.(x*2-3x+3)/(x*2-3x+2)-(x*2-5x+7)/(x*2-5x+6)-1/(x*2-4x+3)2.1/x(x+1)+2/(x+1)(x+3)+3/(x+3)(x+6) 3.已知：x*2-3x+1=0,求x*2+1/x*2. 已知1/√(x)+√(x)=√5,求√(x/x2+x+1)+√(x/x2-x+1)的值定义域 x>0,两边平方，x+2+1/x=5,x+1/x=3,x^2-3x+1=0,x^2+x+1-4x=0,x^2+x+1=4x,x^2-x+1-2x=0.x^2-x+1=2x,√[x/(x^2+x+1)]-√[x/(x^2-x+1)]=√[x/(4x)]-√[x/2x]=√（1/4）-√（1/2） (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 x-1+x+2+x/3+4x=77 先化简,再求值:(2-x/x^2+x)/(x^2-3x/x^2+2x+1)x(9-x^2/4-2x),其中x=-2