恒成立问题,还是判断题.令第一个x式为1式,第二个为2式把两个相加得.再用2式减1式得.明显不符
对任意X(i)存在U(i),V(i),使得X(i)=U(i) -V(i),|X(i)|=U(i)+V(i)
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