(2x+1)/(x-1)
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(2x+1)/(x-1)(2x+1)/(x-1)(2x+1)/(x-1)(2x+1)/(x-1)≤1[(2x+1)/(x-1)]-[(x-1)/(x-1)]≤0(2x+1-x+1)/(x-1)≤0等价
(2x+1)/(x-1)
(2x+1)/(x-1)
(2x+1)/(x-1)
(2x+1)/(x-1)≤1
[(2x+1)/(x-1)]-[(x-1)/(x-1)]≤0
(2x+1-x+1)/(x-1)≤0
等价转换,两个数的商是非正数,那么积也是非正数.
∴(x+2)(x-1)≤0
在数轴上表示（序轴标根法）解得
-2≤x≤1
∵x-1≠0
x≠1
∴这个不等式的解集为{x|-2≤x＜1}
X<=-2
(2x+1)/(x-1)<=1
(2x+1)/(x-1)-1<=0
(2x+1-x+1)/(x-1)<=0
(x+2)(x-1)<=0
-2<=x<1
因为x-1是分母，不等于0所以2x+1小于等于x-1 2x-x小于等于-1-1 x小于等于-2
(2x+1)/(x-1)-1<=0
(2x+1-x+1)/(x-1)<=0
(x+2)/(x-1)<=0
(x+2)(x-1)<=0且x-1不等于0
-2<=x<1
(2x+1)/(x-1)<=1
(2x+1)/(x-1)-1<=0
(2x+1-x+1)/(x-1)<=0
(x+2)/(x-1)<=0
所以-2<=x<=1
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