-x²/(1+x²)
∵1+x²≥1,x²≥0
∴-x²/(1+x²)≤0
∴x≠0时,取任何数分式都是负数
3/(x+2)的值为整数
∴x=1,x=-1,求分式1-X²/(1+xy)²-(x+y)²的值是否能为零,为什么?解(1-x²)/(1+xy)²-(x+y)²(1+x)(1-x)/(1+xy-x-y)(1+xy+x+y)能,当x=1时,或x=-1时,可能为0但是这时取的y值就必须保证分母不能为0说错了,不能为0因为分子为0时,x=1或x=-1这时分母为0
分式-x^2/1+x^2的值为负数,求x的值
分式-x^2/1+x^2的值为负数,求x的值
若3/x+2的值为整数,求x的整数值
若3/x+2的值为整数,求x的整数值
数学人气:131 ℃时间:2019-12-22 16:31:18
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