√[(x-2)/(x-1)]≤1,0≤ (x-2)/(x-1)≤1,0≤1-1/(x-1)≤1,-1≤-1/(x-1)≤0,0≤1/(x-1)≤1,解得x≥2,原不等式的解集为[2,＋∞)；|(x-2)/(x-1)|≤1,-1≤(x-2)/(x-1)≤1,-1≤1-1/(x-1)≤1,-2≤-1/(x-1)≤0,0≤1/(x-1)≤...