2/(x-1)>=(x-1)/(x-1)移项,得(3-x)/(x-1)>=0(x-3)/(x-1)<=0所以 1
把“1”移过来，2/(x-1)-1>=0等价于(3-x)/(x-1)>=0 <=> (3-x)(x-1)>=0 (x?1).故取值范围是1
先移项再解
1
