1. (10 points) The $x$ that does not satisfy the inequality $|x+2|-|x-1|>2$ should meet the condition $\qquad$
1. (10 points) The $x$ that does not satisfy the inequality $|x+2|-|x-1|>2$ should meet the condition $x \leqslant \frac{1}{2}$ -
【Analysis】According to the problem, we need to find the solution to $|x+2|-|x-1| \leqslant 2$. This can be done by discussing three cases: $x1 ;-2 \leqslant x \leqslant 1$, which allows us to remove the absolute values and solve the inequality.
【Solution】Solution: According to the analysis, we need to find the solution to $|x+2|-|x-1| \leqslant 2$. First, we remove the absolute values by considering three cases:
(1) When $x1$, $x+2>0 ; \quad x-1>0$, so: $|x+2|-|x-1| \leqslant 2 \Rightarrow x+2-(x-1) \leqslant 2 \Rightarrow 2+1 \leqslant 2$, (which is not true, discard);
(3) When $-2 \leqslant x \leqslant 1$, $x+2 \geqslant 0 ; x-1 \leqslant 0$, so: $|x+2|-|x-1| \leqslant 2 \Rightarrow x+2-(1-x) \leqslant 2 \Rightarrow 2x+1 \leqslant 2 \Rightarrow x \leqslant \frac{1}{2} ; \quad \therefore-2 \leqslant x \leqslant \frac{1}{2}$.
In summary, when $x \leqslant \frac{1}{2}$, the inequality $|x+2|-|x-1| \leqslant 2$ is satisfied, meaning that at this time, $x$ does not satisfy the inequality $|x+2|-|x-1|>2$. Therefore, the answer is: $x \leqslant \frac{1}{2}$.