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先来看一个貌似简单的数学问题及其解答: 例1 求(x-1)(2~(1/x))≥0的解集. 解由原式得即故所求的解集为:{x|x≥1}. 例2 求x(2~(1/(x-1)))≤0的解集. 解由原式得即故所求的解集为φ。容易看出x=0满足(x-1)(2~(1/x)≥0(即0≥0),故例1所得解集丢掉了x=0这个元素;x=1满足
Let’s look at a seemingly simple mathematical problem and its solution: Example 1 Find the solution set of (x-1)(2~(1/x))≥0. The solution set obtained from the original formula is exactly: {x|x≥1}. Example 2 Find the solution set of x(2~(1/(x-1)))≤0. The solution set obtained from the original formula is exactly φ. It is easy to see that x=0 satisfies (x-1)(2~(1/x)≥0 (that is, 0≥0)), so the solution set in Example 1 loses the x=0 element; x=1 is satisfied.