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x+1\geq 0 x-2<0
For the quotient to be ≤0, one of the values x+1 and x-2 has to be ≥0, the other has to be ≤0, and x-2 cannot be zero. Consider the case when x+1\geq 0 and x-2 is negative.
x\in [-1,2)
滿足兩個不等式的解為 x\in \left[-1,2\right)。
x+1\leq 0 x-2>0
Consider the case when x+1\leq 0 and x-2 is positive.
x\in \emptyset
這對任意 x 均為假。
x\in [-1,2)
最終解是所取得之解的聯集。