lgu 当u>0时单增,由题意,
当x ∈[10,+∞) 时,g(x) = (kx-1)/(x-1) >0,且单增.
(kx-1)/(x-1) = k + (k-1)/(x-1) 单增 => (k-1) < 0 => 0 (kx-1) > 0 => K > 1/10
于是,1/10 < k < 1 .