x^2+(k+1)x+2k-1=0delta=(k+1)^2-4(2k-1)>0k^2+2k+1-8k+4>0k^2-6k+5>0,(k-1)(k-5)>0k属于(-无穷, 1)U(5, +无穷)设两个根为a,b,则-202k-1+(k+1)+1>03k+1>0k属于(-1/3, +无穷)-20(a+2)(b+2)>0ab+2(a+b)+4>02k-1-2(k+1)+4>02k-1-2k-2+4>0, 恒成立k属于(-无穷, 1)U(5, +无穷)k属于(-2, 4)k属于(-1/3, +无穷)综上,k属于(-1/3, 1)
|
