y=(1-x눀)⼀(1+x눀),y的值域?

2025-03-16 16:18:20
推荐回答(2个)
回答1:

因为(1+x²)≥1
所以1/(1+x²)是真分数(或者=1),真分数大于0小于1,
所以得出0<1/(1+x²)≤1这步

因为最后的结果是=-1+2/(1+x²),此处的分子是2,所以下一步的分子是2,懂了没?

y=x-√(x-1) ..........说明x>=1(根号中的数为非负数)
令a=√(x-1)(a>=0) 则x=a*a+1
所以y=a*a+1-a .....................将x用a替换
=(a-1/2)平方+3/4
显然(a-1/2)平方>=0
所以y>=3/4
值域为【3/4,∞)
希望能够帮到你~~

回答2:

为什么会得出0<1/(1+x²)≤1这步呢?
答:把1+x²看做整体k,那么y=1/k(k>=1)反比例函数图象只在第一象限,所以肯定有y>0,因为1+x²大于等于1,所以当分母1+x²比分子1大时,分式的值是不是<1呢?是的

为什么这里的分子是1,而下一步的分子是2?
答:0<1/(1+x²)≤1全式同时乘以2,然后0<2/(1+x²)≤2全式同时-1得-1<-1+2/(1+x²)≤1

y=x-根号x-1 的值域
答:因为根号内x-1≥0,所以x≥1,y≥1

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