已知a是方程x^2-3x+1=0的根,求2a^2-5a-2+2/(a^2+1)的值
1个回答
展开全部
a是方程x^2-3x+1=0的根,
则a^2-3a+1=0;
a^2-2a=a-1;a^2+1=3a.
则2a^2-5a-2+2/(a^2+1)
=2(a^2-2a)-(a+2)+2/(3a)
=2(a-1)-(a+2)+2/(3a)
=a-3+2/(3a)
=[3(a^2+1)-1]/(3a) -3
=(3*3a-1)/(3a) -3
=-1/(3a).
解方程x^2-3x+1=0得:x=a=(3±√5)/2;
则2a^2-5a-2+2/(a^2+1)=-1/(3a)=(-2/3)/(3±√5)
=(3±√5)/6.
则a^2-3a+1=0;
a^2-2a=a-1;a^2+1=3a.
则2a^2-5a-2+2/(a^2+1)
=2(a^2-2a)-(a+2)+2/(3a)
=2(a-1)-(a+2)+2/(3a)
=a-3+2/(3a)
=[3(a^2+1)-1]/(3a) -3
=(3*3a-1)/(3a) -3
=-1/(3a).
解方程x^2-3x+1=0得:x=a=(3±√5)/2;
则2a^2-5a-2+2/(a^2+1)=-1/(3a)=(-2/3)/(3±√5)
=(3±√5)/6.
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询