1÷[(x+1)(x+2)]+1÷[(x+2)(x+3)]+…+1÷[(x+2005)(x+2006)]=1/(2x+4

x(2-1/x)+x/(x^2-2x)÷(3-x)/(x^2-4) x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002 (x+1-1/1-x)÷(x-x²/x-1) (x-4/x-x-6+ x+2/x-3)÷x+1/x-3 x^5+x^4 = (x^3-x)(x^2+x+1)+x^2+x x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4 x-2/x-1÷(x+1-3/x-1) x^2-x/x+1÷x/x+1 ；x^2+6x+9/x^2-9÷x+2/x-3；2/a÷4/a 已知(x-根号2+1)(x-2)=0,求代数式[(x-1)/(x-3)-(x-4)/x]÷（x^2+x-6)/(x^2+ 先化解,在求值;(3x/x-1-x/x+1)÷x四次方-4x²/x³-x,其中x=根号2+2 先化简再求值：（x+2/x²-2x - x-1/x²-4x+4）÷ x-4/x,其中x=3/2 1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...1/(x+2005)(x+2006) 计算:2x/(x-2)(x+1)÷(1－x/x+1)-x²+2x/x²-4?