It is known that 2006 x 2004 x 2005 is found to be 2006 x 2 2004 x 2 values

[(2006-x)-(2004-x)] 22006 and 2004 minus one x each).

2006-x)^2+(2004-x)^2-2(2006-x)(2004-x)

Perfect square formula).

2006-x) 2+(2004-x) 2-2005 2So(2006-x) 2+(2004-x) 2=4+4010=-4014

Because of the slow: 1 + x + x 2 = 0 so x 2007 + x 2006 + x 2005 + ...x+1=1+x+x^2+x^3+x^4+x^5+x6+..

x^2005+x^2006+x^2007=1+x(1+x+x^2)+x^4(1+x+x^2)+.x^2005(1+x+x^2)=1+0+0...0=1 indicates that Zhida imitates Wang:

x+x^2+x^3+x^4+x^5+x6+..x^2005+x^2006+x^200...

1+x+^2+…Both sides of the first match of Qinyin +x 2004+x 2005=0 contain halls multiplied by 1-x

1-x)(1+x+^2+…+x^2004+x^2005)=01-x^2006=0

x^2006=1

Let a=2014-x and b=2012-x, then.

a*b=2013

a-b=2014-x-(2012-x) hand pin=22014-x) 2+(2012-x) 2=a 2+b 2=(a-b) simple circle 2+2ab=2 2+2*2013=4030

Subtract 1 from each fraction to get.

1 (x+2004)+1 (x+2006)=1 (x+2007)+1 (x+2003), and (2x+4010) [(x+2004)(x+2006)]=(2x+4010) [(x+2003)(x+2007)], the denominator on both sides is unequal, 2x+4010=0, x=-2005

It is the root of the original equation.

(2007-x) twice + (2005-x) = (2007-x) twice + (2005-x) - 2(2007-x)(2005-x) + 2(2007-x)(2005-x) =2007-x)-(2005-x)] 2+2*2006 =4+4012=4016