(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0本人有点苯!

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(x+2003)(x+2005)(x+2007)(x+2009)+16=0(x+2003)(x+2005)(x+2007)(x+2009)+16=0本人有点苯!(x+2003)(x+2005)(x+2

(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0本人有点苯!
(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0
(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0
本人有点苯!

(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0本人有点苯!
(x + 2003)(x + 2005)(x + 2007)(x + 2009) + 16 = 0
(x + 2005 -2)(x + 2005)(x + 2007)(x + 2007 +2) + 16 = 0
[(x + 2005)(x + 2007)][(x + 2005 -2)(x + 2007 +2)] + 16 = 0
(前面的变化应该没有什么问题吧?)
不妨设a =(x+2005),b =(x+2007)
于是上式变为
a*b*[(a -2)(b +2)] + 16 = 0
a*b*[a*b -2b +2a -4] +16 =0
将所设的a,b带入,化简,得:
[(x + 2005)(x + 2007)][(x + 2005)(x + 2007) -8] +16 =0
把[(x + 2005)(x + 2007)]看成一项,如果还不明白,不妨再设c =[(x + 2005)(x + 2007)]
上式就变成
c*(c-8) +16 =0
c^2 -8c +16 =0
(c-4)^2 =0
所以,方程的解为:
c-4 =0

[(x + 2005)(x + 2007)] -4 =0
化简,得
x^2 +4012x + 4024031 =0
根据一元二次方程求根公式
应该知道的吧?x =(-b +-SQR((b*b-4ac)/(2a)
所以该方程的根为:
x1 = x3 = (-4012 +SQR(20)/2 = -2006 +SQR(5)
x2 = x4 = (-4012 -SQR(20)/2 = -2006 -SQR(5)
其中,x^2表示x的平方,SQR(5)表示5的平方根.
希望这次能让楼主明白.