由(2x-3)/(x-2)>=1,移项,得:
(2x-3)/(x-2)-1>=0,左边通分,得:
[(2x-3)-(x-2)]/(x-2)>=0,化简,得:
(x-1)/(x-2)>=0.(大于取两边,小于取中间)
所以x2.(分母为x-2,不为0)
(x^3-x^2+x-1)/(x-2)>=0,
[x^2(x-1)+(x-1)]/(x-2)>=0,
(x^2+1)(x-1)/(x-2)>=0,
因为x^2+1>0,恒成立,
所以(x-1)/(x-2)>=0,
所以x2.
从上两不等式的解,可看出:
不等式[(2x-3)/(x-2)]≥1,不等式为[(x³-x²+x-1)/(x-2)]≥0是同解不等式.