已知a是方程x2-2006x+1的根求代数式a2-2005a+2006/a2+1.

a是方程x2-2006x+1的根,a^2-2006a+1=0 a2-2005a+2006/a2+1=2006a-1-2005a+2006/a^2+1=a+2006/(2006a+1) 2.已知实数a.b满足a2-7a+2=0,b2-7b+2=0,说明是方程X2-7X+2=0的两根,满足韦达定理,a+b=7,ab=2 b/a+a/b=(a^2+b^2)/ab=[(a+b)^2-2ab]/ab=(7^2-2*2)/2=22.5