方程x^2+px+2009=0恰有两个正整数根x1,x2,则p/(x1+1)(x2+1)的值是________.
问题描述:
方程x^2+px+2009=0恰有两个正整数根x1,x2,则p/(x1+1)(x2+1)的值是________.
答
x1x2=2009x1+x2=-pp/(x1+1)(x2+1)=p/(x1+x2+x1x2+1)=p/(-p+2009+1)=p/(2010-p)1.2009=1×2009p=-1-2009=-2010p/(x1+1)(x2+1)=-2010/2×2010=-1/22.2009=7×287p=-294p/(x1+1)(x2+1)=-294/(2010+294)=-294/2304=-147/...