xy354710705的个人空间

访问量0

https://bbs.jd100.com/u-62632

k为何值时，不等式-3<[x2+kx+2]/[x2-x+1]</=2对任意实数恒成立

回复(1) 2010-04-18 20:33 来自版块 -

◆ ◆

: teacher2：解答：-3<(x^2+kx+2)/(x^2-x+1) 3+(x^2+kx+2)/(x^2-x+1) =[4x^2+(k-3)x+5]/(x^2-x+1)>0 因为x^2-x+1=(x-1/2)^2+3/4>0 所以4x^2+(k-3)x+5>0 ...(2010-04-18 20:38)

回复

xy354710705

加关注写私信打招呼